case study questions from trigonometry class 10

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CBSE Class 10 Maths Case Study Questions for Chapter 9 - Some Applications of Trigonometry (Published By CBSE)

Check case study questions for cbse class 10 maths chapter 9 - some applications of trigonometry. these questions are published by the cbse itself for class 10 students..

Case study based questions are new for class 10 students. Therefore, it is quite essential that students practice with more of such questions so that they do not have problem in solving them in their Maths board exam. We have provided here the case study questions for CBSE Class 10 Maths Chapter 9 - Some Applications of Trigonometry. All these questions have been published by the Central Board of Secondary Education (CBSE) for the class 10 students. Therefore, students must solve all the questions seriously so that they may score the desired marks in their Maths exam.

Check Case Study Questions for Class 10 Maths Chapter 9:

CASE STUDY 1:

A group of students of class X visited India Gate on an education trip. The teacher and students had interest in history as well. The teacher narrated that India Gate, official name Delhi Memorial, originally called All-India War Memorial, monumental sandstone arch in New Delhi, dedicated to the troops of British India who died in wars fought between 1914 and 1919. The teacher also said that India Gate, which is located at the eastern end of the Rajpath (formerly called the Kingsway), is about 138 feet (42 metres) in height.

1. What is the angle of elevation if they are standing at a distance of 42m away from the monument?

Answer: b) 45 o

2. They want to see the tower at an angle of 60 o . So, they want to know the distance where they should stand and hence find the distance.

Answer: a) 25.24 m

3. If the altitude of the Sun is at 60 o , then the height of the vertical tower that will cast a shadow of length 20 m is

a) 20√3 m

b) 20/ √3 m

c) 15/ √3 m

d) 15√3 m

Answer: a) 20√3 m

4. The ratio of the length of a rod and its shadow is 1:1. The angle of elevation of the Sun is

5. The angle formed by the line of sight with the horizontal when the object viewed is below the horizontal level is

a) corresponding angle

b) angle of elevation

c) angle of depression

d) complete angle

Answer: a) corresponding angle

CASE STUDY 2:

A Satellite flying at height h is watching the top of the two tallest mountains in Uttarakhand and Karnataka, them being Nanda Devi(height 7,816m) and Mullayanagiri (height 1,930 m). The angles of depression from the satellite, to the top of Nanda Devi and Mullayanagiri are 30° and 60° respectively. If the distance between the peaks of the two mountains is 1937 km, and the satellite is vertically above the midpoint of the distance between the two mountains.

1. The distance of the satellite from the top of Nanda Devi is

a) 1139.4 km

b) 577.52 km

d) 1025.36 km

Answer: a) 1139.4 km

2. The distance of the satellite from the top of Mullayanagiri is

Answer: c) 1937 km

3. The distance of the satellite from the ground is

Answer: b) 577.52 km

4. What is the angle of elevation if a man is standing at a distance of 7816m from Nanda Devi?

5.If a mile stone very far away from, makes 45 o to the top of Mullanyangiri mountain. So, find the distance of this mile stone from the mountain.

a) 1118.327 km

b) 566.976 km

Also Check:

Case Study Questions for All Chapters of CBSE Class 10 Maths

Tips to Solve Case Study Based Questions Accurately

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Class 10 Maths Case Study Questions Chapter 8 Introduction to Trigonometry

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Case study Questions in the Class 10 Mathematics Chapter 8  are very important to solve for your exam. Class 10 Maths Chapter 8 Case Study Questions have been prepared for the latest exam pattern. You can check your knowledge by solving Class 10 Maths Case Study Questions Chapter 8  Introduction to Trigonometry

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In CBSE Class 10 Maths Paper, Students will have to answer some questions based on Assertion and Reason. There will be a few questions based on case studies and passage-based as well. In that, a paragraph will be given, and then the MCQ questions based on it will be asked.

Introduction to Trigonometry Case Study Questions With Answers

Here, we have provided case-based/passage-based questions for Class 10 Maths  Chapter 8 Introduction to Trigonometry

Case Study/Passage-Based Questions

Question 1:

Answer: (d) 6m

(ii) Measure of ∠A =

Answer: (c) 45°

(iii) Measure of ∠C =

(iv) Find the value of sinA + cosC.

Answer: (d) 2√2

(v) Find the value of tan 2 C + tan 2  A.

Answer: (c) 2

Question 2:

Answer: (a) 30°

(ii) The measure of  ∠C is

Answer: (c) 60°

(iii) The length of AC is 

Answer: (d)6√3m

(iv) cos2A =

Answer: (b)1/2

Hope the information shed above regarding Case Study and Passage Based Questions for Class 10 Maths Chapter 8 Introduction to Trigonometry with Answers Pdf free download has been useful to an extent. If you have any other queries about CBSE Class 10 Maths Introduction to Trigonometry Case Study and Passage Based Questions with Answers, feel free to comment below so that we can revert back to us at the earliest possible By Team Study Rate

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CBSE Case Study Questions for Class 10 Maths Trigonometry Free PDF

Mere Bacchon, you must practice the CBSE Case Study Questions Class 10 Maths Trigonometry  in order to fully complete your preparation . They are very very important from exam point of view. These tricky Case Study Based Questions can act as a villain in your heroic exams!

I have made sure the questions (along with the solutions) prepare you fully for the upcoming exams. To download the latest CBSE Case Study Questions , just click ‘ Download PDF ’.

CBSE Case Study Questions for Class 10 Maths Trigonometry PDF

Checkout our case study questions for other chapters.

• Chapter 6 Triangles Case Study Questions
• Chapter 7 Coordinate Geometry Case Study Questions
• Chapter 9 Some Applications of Trigonometry Case Study Questions
• Chapter 10 Circles Case Study Questions

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Chapter 8 Class 10 Introduction to Trignometry

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The chapter is updated according to the new NCERT, for 2023-2024 Board Exams.

Get NCERT Solutions with videos of all questions and examples of Chapter 8 Class 10 Trigonometry. Videos of all questions are made with step-by-step explanations. Check it out now.

Trigonometry means studying relationship between measures of triangle. Usually, we talk about right triangles when we study trigonometry,

In this chapter, we will study

• What is sin, cos, tan ( Sine, cosine, tangent) ... and how they are found in a triangle
• What is sec, cosec, cot, and how is it related to sin, cos, tan.
• (Sin, cos, tan, sec, cosec, cot are known as Trigonometric Ratios)
• Then, we study Trigonometric ratios of specific angles l ike 0°, 30°, 45°, 60°, 90° ; and do some questions
• We study the formulas of sin (90 - θ) , cos (90 - θ), tan (90 - θ)
• And then we study Trigonometric Identities, and how other identities are derived from sin 2 θ + cos 2 θ = 1

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CBSE Class 10 Maths Case Study Questions PDF

Download Case Study Questions for Class 10 Mathematics to prepare for the upcoming CBSE Class 10 Final Exam. These Case Study and Passage Based questions are published by the experts of CBSE Experts for the students of CBSE Class 10 so that they can score 100% on Boards.

CBSE Class 10 Mathematics Exam 2024  will have a set of questions based on case studies in the form of MCQs. The CBSE Class 10 Mathematics Question Bank on Case Studies, provided in this article, can be very helpful to understand the new format of questions. Share this link with your friends.

Table of Contents

Chapterwise Case Study Questions for Class 10 Mathematics

Inboard exams, students will find the questions based on assertion and reasoning. Also, there will be a few questions based on case studies. In that, a paragraph will be given, and then the MCQ questions based on it will be asked.

The above  Case studies for Class 10 Maths will help you to boost your scores as Case Study questions have been coming in your examinations. These CBSE Class 10 Mathematics Case Studies have been developed by experienced teachers of cbseexpert.com for the benefit of Class 10 students.

• Class 10th Science Case Study Questions
• Assertion and Reason Questions of Class 10th Science
• Assertion and Reason Questions of Class 10th Social Science

Class 10 Maths Syllabus 2024

Chapter-1  real numbers.

Starting with an introduction to real numbers, properties of real numbers, Euclid’s division lemma, fundamentals of arithmetic, Euclid’s division algorithm, revisiting irrational numbers, revisiting rational numbers and their decimal expansions followed by a bunch of problems for a thorough and better understanding.

Chapter-2  Polynomials

This chapter is quite important and marks securing topics in the syllabus. As this chapter is repeated almost every year, students find this a very easy and simple subject to understand. Topics like the geometrical meaning of the zeroes of a polynomial, the relationship between zeroes and coefficients of a polynomial, division algorithm for polynomials followed with exercises and solved examples for thorough understanding.

Chapter-3  Pair of Linear Equations in Two Variables

This chapter is very intriguing and the topics covered here are explained very clearly and perfectly using examples and exercises for each topic. Starting with the introduction, pair of linear equations in two variables, graphical method of solution of a pair of linear equations, algebraic methods of solving a pair of linear equations, substitution method, elimination method, cross-multiplication method, equations reducible to a pair of linear equations in two variables, etc are a few topics that are discussed in this chapter.

Chapter-4  Quadratic Equations

The Quadratic Equations chapter is a very important and high priority subject in terms of examination, and securing as well as the problems are very simple and easy. Problems like finding the value of X from a given equation, comparing and solving two equations to find X, Y values, proving the given equation is quadratic or not by knowing the highest power, from the given statement deriving the required quadratic equation, etc are few topics covered in this chapter and also an ample set of problems are provided for better practice purposes.

Chapter-5  Arithmetic Progressions

This chapter is another interesting and simpler topic where the problems here are mostly based on a single formula and the rest are derivations of the original one. Beginning with a basic brief introduction, definitions of arithmetic progressions, nth term of an AP, the sum of first n terms of an AP are a few important and priority topics covered under this chapter. Apart from that, there are many problems and exercises followed with each topic for good understanding.

Chapter-6  Triangles

This chapter Triangle is an interesting and easy chapter and students often like this very much and a securing unit as well. Here beginning with the introduction to triangles followed by other topics like similar figures, the similarity of triangles, criteria for similarity of triangles, areas of similar triangles, Pythagoras theorem, along with a page summary for revision purposes are discussed in this chapter with examples and exercises for practice purposes.

Chapter-7  Coordinate Geometry

Here starting with a general introduction, distance formula, section formula, area of the triangle are a few topics covered in this chapter followed with examples and exercises for better and thorough practice purposes.

Chapter-8  Introduction to Trigonometry

As trigonometry is a very important and vast subject, this topic is divided into two parts where one chapter is Introduction to Trigonometry and another part is Applications of Trigonometry. This Introduction to Trigonometry chapter is started with a general introduction, trigonometric ratios, trigonometric ratios of some specific angles, trigonometric ratios of complementary angles, trigonometric identities, etc are a few important topics covered in this chapter.

Chapter-9  Applications of Trigonometry

This chapter is the continuation of the previous chapter, where the various modeled applications are discussed here with examples and exercises for better understanding. Topics like heights and distances are covered here and at the end, a summary is provided with all the important and frequently used formulas used in this chapter for solving the problems.

Chapter-10  Circle

Beginning with the introduction to circles, tangent to a circle, several tangents from a point on a circle are some of the important topics covered in this chapter. This chapter being practical, there are an ample number of problems and solved examples for better understanding and practice purposes.

Chapter-11  Constructions

This chapter has more practical problems than theory-based definitions. Beginning with a general introduction to constructions, tools used, etc, the topics like division of a line segment, construction of tangents to a circle, and followed with few solved examples that help in solving the exercises provided after each topic.

Chapter-12  Areas related to Circles

This chapter problem is exclusively formula based wherein topics like perimeter and area of a circle- A Review, areas of sector and segment of a circle, areas of combinations of plane figures, and a page summary is provided just as a revision of the topics and formulas covered in the entire chapter and also there are many exercises and solved examples for practice purposes.

Chapter-13  Surface Areas and Volumes

Starting with the introduction, the surface area of a combination of solids, the volume of a combination of solids, conversion of solid from one shape to another, frustum of a cone, etc are to name a few topics explained in detail provided with a set of examples for a better comprehension of the concepts.

Chapter-14  Statistics

In this chapter starting with an introduction, topics like mean of grouped data, mode of grouped data, a median of grouped, graphical representation of cumulative frequency distribution are explained in detail with exercises for practice purposes. This chapter being a simple and easy subject, securing the marks is not difficult for students.

Chapter-15  Probability

Probability is another simple and important chapter in examination point of view and as seeking knowledge purposes as well. Beginning with an introduction to probability, an important topic called A theoretical approach is explained here. Since this chapter is one of the smallest in the syllabus and problems are also quite easy, students often like this chapter

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• Class 10 Maths
• Chapter 8: Introduction Trigonometry

Important Questions for Class 10 Maths Chapter 8- Introduction to Trigonometry

Important questions of Chapter 8 – Introduction to Trigonometry of Class 10 are given here for students who want to score high marks in their board exams 2022-2023. Students can also download the trigonometry class 10 questions pdf which is available on our website. They can learn and solve the problems offline by downloading trigonometry class 10 questions pdf.  The questions which are provided below are as per the latest CBSE syllabus and are designed according to the NCERT book. The questions are formulated after analyzing the previous year’s questions papers, exam trends and latest sample papers. Solving these questions will help students to get prepared for the final exam. Students can also get important questions for all the chapters of 10th standard Maths . Solve them to get acquainted with the various types of questions to be asked from each chapter of the Maths subject.

Also, get access to Class 10 Maths Chapter 8 Introduction to Trigonometry MCQs here.

In Chapter 8, students will be introduced to the Trigonometry concept, which states the relationship between angles and sides of a triangle. They will come across many trigonometric formulas which will be used to solve numerical problems. The six primary trigonometric ratios are sine, cosine, tangent, secant, cosecant and cotangent. The whole trigonometry concept revolves around these ratios or sometimes also called functions. Students can learn more about trigonometry class 10 questions, which are provided with complete explanations. Trigonometry Class 10 Maths Chapter 8 Important Questions are solved by our expert teachers so that students can understand the problems quickly. Besides, students can also get additional questions on chapter 8 of class 10 maths for practice at the end.

Class 10 Maths Chapter 8 Important Questions and Answers

Below are the important trigonometry class 10 questions. Students can refer to the below-given class 10 trigonometry questions and they can practice these problems as well.

Question. 1 : In ∆ ABC, right-angled at B, AB = 24 cm, BC = 7 cm.

Determine: (i) sin A, cos A (ii) sin C, cos C

In a given triangle ABC, right-angled at B = ∠B = 90°

Given: AB = 24 cm and BC = 7 cm

That means, AC = Hypotenuse

According to the Pythagoras Theorem,

In a right-angled triangle, the squares of the hypotenuse side are equal to the sum of the squares of the other two sides.

By applying the Pythagoras theorem, we get

AC 2  = AB 2  + BC 2

AC 2  = (24) 2  + 7 2

AC 2  = (576 + 49)

AC 2  = 625 cm 2

Therefore, AC = 25 cm

(i) We need to find Sin A and Cos A.

As we know, sine of the angle is equal to the ratio of the length of the opposite side and hypotenuse side. Therefore,

Sin A = BC/AC = 7/25

Again, the cosine of an angle is equal to the ratio of the adjacent side and hypotenuse side. Therefore,

cos A = AB/AC = 24/25

(ii) We need to find Sin C and Cos C.

Sin C = AB/AC = 24/25

Cos C = BC/AC = 7/25

Question 2: If Sin A = 3/4, Calculate cos A and tan A.

Solution: Let us say, ABC is a right-angled triangle, right-angled at B.

Sin A = 3/4

As we know,

Sin A = Opposite Side/Hypotenuse Side = 3/4

Now, let BC be 3k and AC will be 4k.

where k is the positive real number.

As per the Pythagoras theorem, we know;

Hypotenuse 2 = Perpendicular 2 + Base 2

AC 2 = AB 2 + BC 2

Substitute the value of AC and BC in the above expression to get;

(4k) 2 = (AB) 2 + (3k) 2

16k 2 – 9k 2 = AB 2

AB 2 = 7k 2

Hence, AB =  √7 k 

Now, as per the question, we need to find the value of cos A and tan A.

cos A = Adjacent Side/Hypotenuse side = AB/AC

cos A =  √7 k/4k = √7/4

tan A = Opposite side/Adjacent side = BC/AB

tan A = 3k/ √7 k = 3/√7

Question.3: If ∠A and ∠B are acute angles such that cos A = cos B, then show that ∠ A = ∠ B.

Suppose a triangle ABC, right-angled at C.

Now, we know the trigonometric ratios,

cos A = AC/AB

cos B = BC/AB

Since, it is given,

cos A = cos B

AC/AB = BC/AB

We know that by isosceles triangle theorem, the angles opposite to the equal sides are equal.

Therefore, ∠A = ∠B

Question 4: If 3 cot A = 4, check whether (1 – tan 2 A)/(1 + tan 2 A) = cos 2 A – sin 2 A or not.

Let us consider a triangle ABC, right-angled at B.

3 cot A = 4

cot A = 4/3

Since, tan A = 1/cot A

tan A = 1/(4/3) = 3/4

BC/AB = 3/4

Let BC = 3k and AB = 4k

By using Pythagoras theorem, we get;

AC 2 = (4k) 2 + (3k) 2

AC 2 = 16k 2  + 9k 2

AC = √ 25k 2  = 5k

sin A = Opposite side/Hypotenuse

In the same way,

cos A = Adjacent side/hypotenuse

To check: (1-tan 2 A)/(1+tan 2 A) = cos 2 A – sin 2 A or not

Let us take L.H.S. first;

= [1 – (9/16)]/[1 + (9/16)] = 7/25

R.H.S. = cos 2 A – sin 2 A = (4/5) 2  – (3/5) 2

= (16/25) – (9/25) = 7/25

L.H.S. = R.H.S.

Hence, proved.

Question 5: In triangle PQR, right-angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values of sin P, cos P and tan P.

Solution: Given,

In triangle PQR,

PR + QR = 25 cm

Let us say, QR = x

Then, PR = 25 – QR = 25 – x

Using Pythagoras theorem:

PR 2 = PQ 2 + QR 2

Now, substituting the value of PR, PQ and QR, we get;

(25 – x) 2 = (5) 2 + (x) 2

25 2  + x 2  – 50x = 25 + x 2

625 – 50x = 25

So, QR = 12 cm

PR = 25 – QR = 25 – 12 = 13 cm

sin P = QR/PR = 12/13

cos P = PQ/PR = 5/13

tan P = QR/PQ = 12/5

Question 6: Evaluate 2 tan 2 45° + cos 2 30° – sin 2 60°.

Solution: Since we know,

tan 45° = 1

cos 30° = √ 3/2

sin 60° = √ 3/2

Therefore, putting these values in the given equation:

2(1) 2 + ( √ 3/2) 2 – ( √ 3/2) 2

Question 7: If tan (A + B) =√3 and tan (A – B) =1/√3, 0° < A + B ≤ 90°; A > B, find A and B.

tan (A + B) = √3

As we know, tan 60° = √3

Thus, we can write;

⇒ tan (A + B) = tan 60°

⇒(A + B) = 60° …… (i)

Now again given;

tan (A – B) = 1/√3

Since, tan 30° = 1/√3

⇒ tan (A – B) = tan 30°

⇒(A – B) = 30° ….. (ii)

Adding the equation (i) and (ii), we get;

A + B + A – B = 60° + 30°

Now, put the value of A in eq. (i) to find the value of B;

45° + B = 60°

B = 60° – 45°

Therefore A = 45° and B = 15°

Question 8: Show that :

(i) tan 48° tan 23° tan 42° tan 67° = 1

(ii) cos 38° cos 52° – sin 38° sin 52° = 0

(i) tan 48° tan 23° tan 42° tan 67°

We can also write the above given tan functions in terms of cot functions, such as;

tan 48° = tan (90° – 42°) = cot 42°

tan 23° = tan (90° – 67°) = cot 67°

Hence, substituting these values, we get

= cot 42° cot 67° tan 42° tan 67°

= (cot 42° tan 42°) (cot 67° tan 67°)

(ii) cos 38° cos 52° – sin 38° sin 52°

We can also write the given cos functions in terms of sin functions.

cos 38° = cos (90° – 52°) = sin 52°

cos 52°= cos (90° – 38°) = sin 38°

Hence, putting these values in the given equation, we get;

sin 52° sin 38° – sin 38° sin 52° = 0

Question 9: If tan 2A = cot (A – 18°), where 2A is an acute angle, find the value of A.

tan 2A = cot (A – 18°)

As we know by trigonometric identities,

tan 2A = cot (90° – 2A)

Substituting the above equation in the given equation, we get;

⇒ cot (90° – 2A) = cot (A – 18°)

⇒ 90° – 2A = A – 18°

⇒ 108° = 3A

A = 108° / 3

Hence, the value of A = 36°

Question 10:  If A, B and C are interior angles of a triangle ABC, then show that sin [(B + C)/2] = cos A/2.

As we know, for any given triangle, the sum of all its interior angles is equals to 180°.

A + B + C = 180° ….(1)

Now we can write the above equation as;

⇒ B + C = 180° – A

Dividing by 2 on both the sides;

⇒ (B + C)/2 = (180° – A)/2

⇒ (B + C)/2 = 90° – A/2

Now, put sin function on both sides.

⇒ sin (B + C)/2 = sin (90° – A/2)

sin (90° – A/2) = cos A/2

sin (B + C)/2 = cos A/2

Question 11: Prove the identities:

(i) √[1 + sinA/1 – sinA] = sec A + tan A

(ii) (1 + tan 2 A/1 + cot 2 A) = (1 – tan A/1 – cot A) 2 = tan 2 A Solution: (i) Given:√[1 + sinA/1 – sinA] = sec A + tan A

LHS: = (1+tan²A) / (1+cot²A) Using the trigonometric identities we know that 1+tan²A = sec²A and 1+cot²A= cosec²A = sec²A/ cosec²A On taking the reciprocals we get = sin²A/cos²A = tan²A RHS: =(1-tanA)²/(1-cotA)² Substituting the reciprocal value of tan A and cot A we get, = (1-sinA/cosA)²/(1-cosA/sinA)² = [(cosA-sinA)/cosA]²/ [(sinA-cos)/sinA)²] = [(cosA-sinA)²×sin²A] /[cos²A. /(sinA-cosA)²] =  sin²A/cos²A = tan 2 A The values of LHS and RHS are the same. Hence proved.

Question 12: If sin θ + cos θ = √3, then prove that tan θ + cot θ = 1.

sin θ + cos θ = √3

Squaring on both sides,

(sin θ + cos θ) 2 = (√3) 2

sin 2 θ + cos 2 θ + 2 sin θ cos θ = 3

Using the identity sin 2 A + cos 2 A = 1,

1 + 2 sin θ cos θ = 3

2 sin θ cos θ = 3 – 1

2 sin θ cos θ = 2

sin θ cos θ = 1

sin θ cos θ = sin 2 θ + cos 2 θ

⇒ (sin 2 θ + cos 2 θ)/(sin θ cos θ) = 1

⇒ [sin 2 θ/(sin θ cos θ)] + [cos 2 θ/(sin θ cos θ)] = 1

⇒ (sin θ/cos θ) + (cos θ/sin θ) = 1

⇒ tan θ + cot θ = 1

Hence proved.

Question 13: Express cot 85° + cos 75° in terms of trigonometric ratios of angles between 0° and 45°.

cot 85° + cos 75° 

= cot (90° – 5°) + cos (90° – 15°) 

We know that cos(90° – A) = sin A and cot(90° – A) = tan A

= tan 5° + sin 15°

Question 14: What is the value of (cos 2 67° – sin 2 23°)?

(cos 2 67° – sin 2 23°) = cos 2 (90° – 23°) – sin 2 23°

We know that cos(90° – A) = sin A

= sin 2 23° – sin 2 23°

Therefore, (cos 2 67° – sin 2 23°) = 1.

Question 15: Prove that (sin A – 2 sin 3 A)/(2 cos 3 A – cos A) = tan A.

LHS = (sin A – 2 sin 3 A)/(2 cos 3 A – cos A)

= [sin A(1 – 2 sin 2 A)]/ [cos A(2 cos 2 A – 1]

Using the identity sin 2 θ + cos 2 θ = 1,

= [sin A(sin 2 A + cos 2 A – 2 sin 2 A)]/ [cos A(2 cos 2 A – sin 2 A – cos 2 A]

= [sin A(cos 2 A – sin 2 A)]/ [cos A(cos 2 A – sin 2 A)]

= sin A/cos A

Also, check: Trigonometry Class 10 Questions PDF .

Video Lesson on Trigonometry

Introduction to Trigonometry Class 10 Questions for Practice

Solve the following class 10 Maths trigonometry problems:

• If sec θ + tan θ = 7, the compute the value of sec θ – tan θ
• If tan θ + cot θ = 5, then find the value of tan 2 θ + cot 2 θ.
• What will happen if the value of the cosine function increases from 0° to 90°?
• Given that cosec θ = 4/3. Find all other values of trigonometric ratios.
• Find the value of sin 45 °- Cos 45°.
• If 4 tan θ = 3, evaluate (4 sin θ – cos θ + 1)/(4 sin θ + cos θ + 1).
• Prove that [tan θ/(1 – cot θ)] + [cot θ/(1 – tan θ)] = 1 + sec θ cosec θ
• Prove that: sin θ/(cot θ + cosec θ) = 2 + [sin θ/ (cot θ – cosec θ]
• In a right triangle ABC, right-angled at B, if tan A = 1, then verify that 2 sin A cos A = 1.
• If ∠B and ∠Q are acute angles such that sin B = sin Q, then prove that ∠B = ∠Q.

We hope students must have found this information on Important Questions Class 10 Maths Chapter 8 Introduction to Trigonometry helpful for their board exam preparation. Stay tuned with BYJU’S – The Learning App and download the app to get all the chapter-wise important questions for class 10 Maths and also get the latest updates on the syllabus.

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Nice work by byjus ..

It is very good work from byjus and this questions are very helpful for us

Good short and easy but important one’s

thank u soo much byju

These are easy, please upload most important and 100% asking questions in board

That’s very helpful to me during my exams. Thanks to BYJU’S App.

Thank you byjus. I was searching for trigonometry related this type questions only.

Thank you Byju’s the learning app,this is going to be much helpful for us Please upload more questions similar to this with a little bit more complexity. By the way, I am one of those Byju’s offline users, but yet searching in google for more questions Please upload the most important EXERCISE QUESTIONS and EXAMPLE QUESTIONS from textbook as well

Thank you Byju’s this helped me a lot in my internal assessments.

Thank you, BYJU’S ! It helped me a lot for my PRE-BOARD.

If cosec+ cot=k them prove that cos=k²-1/k²+1

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• Case Study Class 10...

Case Study Class 10 Maths Questions

Table of Contents

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Download the app to get CBSE Sample Papers 2023-24, NCERT Solutions (Revised), Most Important Questions, Previous Year Question Bank, Mock Tests, and Detailed Notes.

Now, CBSE will ask only subjective questions in class 10 Maths case studies. But if you search over the internet or even check many books, you will get only MCQs in the class 10 Maths case study in the session 2022-23. It is not the correct pattern. Just beware of such misleading websites and books.

We advise you to visit CBSE official website ( cbseacademic.nic.in ) and go through class 10 model question papers . You will find that CBSE is asking only subjective questions under case study in class 10 Maths. We at myCBSEguide helping CBSE students for the past 15 years and are committed to providing the most authentic study material to our students.

Here, myCBSEguide is the only application that has the most relevant and updated study material for CBSE students as per the official curriculum document 2022 – 2023. You can download updated sample papers for class 10 maths .

First of all, we would like to clarify that class 10 maths case study questions are subjective and CBSE will not ask multiple-choice questions in case studies. So, you must download the myCBSEguide app to get updated model question papers having new pattern subjective case study questions for class 10 the mathematics year 2022-23.

Class 10 Maths has the following chapters.

• Real Numbers Case Study Question
• Polynomials Case Study Question
• Pair of Linear Equations in Two Variables Case Study Question
• Quadratic Equations Case Study Question
• Arithmetic Progressions Case Study Question
• Triangles Case Study Question
• Coordinate Geometry Case Study Question
• Introduction to Trigonometry Case Study Question
• Some Applications of Trigonometry Case Study Question
• Circles Case Study Question
• Area Related to Circles Case Study Question
• Surface Areas and Volumes Case Study Question
• Statistics Case Study Question
• Probability Case Study Question

Format of Maths Case-Based Questions

CBSE Class 10 Maths Case Study Questions will have one passage and four questions. As you know, CBSE has introduced Case Study Questions in class 10 and class 12 this year, the annual examination will have case-based questions in almost all major subjects. This article will help you to find sample questions based on case studies and model question papers for CBSE class 10 Board Exams.

Maths Case Study Question Paper 2023

Here is the marks distribution of the CBSE class 10 maths board exam question paper. CBSE may ask case study questions from any of the following chapters. However, Mensuration, statistics, probability and Algebra are some important chapters in this regard.

Case Study Question in Mathematics

Here are some examples of case study-based questions for class 10 Mathematics. To get more questions and model question papers for the 2021 examination, download myCBSEguide Mobile App .

Case Study Question – 1

In the month of April to June 2022, the exports of passenger cars from India increased by 26% in the corresponding quarter of 2021–22, as per a report. A car manufacturing company planned to produce 1800 cars in 4th year and 2600 cars in 8th year. Assuming that the production increases uniformly by a fixed number every year.

• Find the production in the 1 st year.
• Find the production in the 12 th year.
• Find the total production in first 10 years. OR In which year the total production will reach to 15000 cars?

Case Study Question – 2

In a GPS, The lines that run east-west are known as lines of latitude, and the lines running north-south are known as lines of longitude. The latitude and the longitude of a place are its coordinates and the distance formula is used to find the distance between two places. The distance between two parallel lines is approximately 150 km. A family from Uttar Pradesh planned a round trip from Lucknow (L) to Puri (P) via Bhuj (B) and Nashik (N) as shown in the given figure below.

• Find the distance between Lucknow (L) to Bhuj(B).
• If Kota (K), internally divide the line segment joining Lucknow (L) to Bhuj (B) into 3 : 2 then find the coordinate of Kota (K).
• Name the type of triangle formed by the places Lucknow (L), Nashik (N) and Puri (P) OR Find a place (point) on the longitude (y-axis) which is equidistant from the points Lucknow (L) and Puri (P).

Case Study Question – 3

• Find the distance PA.
• Find the distance PB
• Find the width AB of the river. OR Find the height BQ if the angle of the elevation from P to Q be 30 o .

Case Study Question – 4

• What is the length of the line segment joining points B and F?
• The centre ‘Z’ of the figure will be the point of intersection of the diagonals of quadrilateral WXOP. Then what are the coordinates of Z?
• What are the coordinates of the point on y axis equidistant from A and G? OR What is the area of area of Trapezium AFGH?

Case Study Question – 5

The school auditorium was to be constructed to accommodate at least 1500 people. The chairs are to be placed in concentric circular arrangement in such a way that each succeeding circular row has 10 seats more than the previous one.

• If the first circular row has 30 seats, how many seats will be there in the 10th row?
• For 1500 seats in the auditorium, how many rows need to be there? OR If 1500 seats are to be arranged in the auditorium, how many seats are still left to be put after 10 th row?
• If there were 17 rows in the auditorium, how many seats will be there in the middle row?

Case Study Question – 6

• Draw a neat labelled figure to show the above situation diagrammatically.

• What is the speed of the plane in km/hr.

More Case Study Questions

We have class 10 maths case study questions in every chapter. You can download them as PDFs from the myCBSEguide App or from our free student dashboard .

As you know CBSE has reduced the syllabus this year, you should be careful while downloading these case study questions from the internet. You may get outdated or irrelevant questions there. It will not only be a waste of time but also lead to confusion.

Here, myCBSEguide is the most authentic learning app for CBSE students that is providing you up to date study material. You can download the myCBSEguide app and get access to 100+ case study questions for class 10 Maths.

How to Solve Case-Based Questions?

Questions based on a given case study are normally taken from real-life situations. These are certainly related to the concepts provided in the textbook but the plot of the question is always based on a day-to-day life problem. There will be all subjective-type questions in the case study. You should answer the case-based questions to the point.

What are Class 10 competency-based questions?

Competency-based questions are questions that are based on real-life situations. Case study questions are a type of competency-based questions. There may be multiple ways to assess the competencies. The case study is assumed to be one of the best methods to evaluate competencies. In class 10 maths, you will find 1-2 case study questions. We advise you to read the passage carefully before answering the questions.

Case Study Questions in Maths Question Paper

CBSE has released new model question papers for annual examinations. myCBSEguide App has also created many model papers based on the new format (reduced syllabus) for the current session and uploaded them to myCBSEguide App. We advise all the students to download the myCBSEguide app and practice case study questions for class 10 maths as much as possible.

Case Studies on CBSE’s Official Website

CBSE has uploaded many case study questions on class 10 maths. You can download them from CBSE Official Website for free. Here you will find around 40-50 case study questions in PDF format for CBSE 10th class.

10 Maths Case Studies in myCBSEguide App

You can also download chapter-wise case study questions for class 10 maths from the myCBSEguide app. These class 10 case-based questions are prepared by our team of expert teachers. We have kept the new reduced syllabus in mind while creating these case-based questions. So, you will get the updated questions only.

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Related Posts

• CBSE Class 10 Maths Sample Paper 2020-21
• Class 12 Maths Case Study Questions
• CBSE Reduced Syllabus Class 10 (2020-21)
• Class 10 Maths Basic Sample Paper 2024
• How to Revise CBSE Class 10 Maths in 3 Days
• CBSE Practice Papers 2023
• Class 10 Maths Sample Papers 2024
• Competency Based Learning in CBSE Schools

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CBSE 10th Standard Maths Subject Some Applications of Trigonometry Case Study Questions 2021

By QB365 on 22 May, 2021

QB365 Provides the updated CASE Study Questions for Class 10 Maths, and also provide the detail solution for each and every case study questions . Case study questions are latest updated question pattern from NCERT, QB365 will helps to get  more marks in Exams

QB365 - Question Bank Software

10th Standard CBSE

Final Semester - June 2015

Case Study Questions

(ii) Measure of \(\angle\) ACB is equal to

(iii) Width of the river is 

(iv) Height of the other temple is

(v) Angle of depression is always

(ii) Value of DF is equal to

(iii) Value of h is

(iv) Height of the balloon from the ground is

(v) If the balloon is moving towards the building, then both angle of elevation will

(ii) If the angle made by the rope to the ground level is 45°, then find the distance between artist and pole at ground level.

(iii) Find the height of the pole if the angle made by the rope to the ground level is 30°.

(iv) If the angle made by the rope to the ground level is 30° and 3 m rope is broken, then find the height of the pole

(v) Which mathematical concept is used here?

(ii) If fireman place the ladder 5 m away from the wall and angle of elevation is observed to be 30°, then length of the ladder is

(iii) If fireman places the ladder 2.5 m away from the wall and angle of elevation is observed to be 60°, then find the height of the window. (Take \(\sqrt{3}\) = 1.73)

(iv) If the height of the window is 8 m above the ground and angle of elevation is observed to be 45°, then horizontal distance between the foot of ladder and wall is

(v) If the fireman gets a 9 m long ladder and window is at 6 m height, then how far should the ladder be placed?

(ii) Distance between two positions of the car is

(iii) Total time taken by the car to reach the foot of the building from starting point is

(iv) The distance of the observer from the car when it makes an angle of 60° is

(v) The angle of elevation increases

*****************************************

Cbse 10th standard maths subject some applications of trigonometry case study questions 2021 answer keys.

(i) (b): The person who makes small angle of elevation is more closer to the balloon. \(\therefore\) Radlra is more closer to the balloon. (ii) (b):  \(\text { In } \Delta E F D, \tan 30^{\circ}=\frac{E D}{D F}\) \(\Rightarrow \quad \frac{1}{\sqrt{3}}=\frac{h}{D F} \) \(\Rightarrow \quad D F=h \sqrt{3} \mathrm{~m}\) (iii) (a): In \(\Delta\) GCE, \(\begin{array}{l} \tan 60^{\circ}=\frac{E C}{G C}=\frac{h+4}{D F} \\ \Rightarrow \quad \sqrt{3}=\frac{h+4}{\sqrt{3} h} \Rightarrow 3 h=h+4 \Rightarrow h=2 \end{array}\) (iv) (c): Height of the balloon from the ground = BE = BC + CD + DE = 2 + 4 + 2 = 8 m (v) (b)

(i) (c):   \(\text { In } \Delta A B C, \frac{A B}{B C}=\tan 60^{\circ}\) \(\Rightarrow \quad A B=25 \times \sqrt{3}\) \(\therefore\) Height of building is 25 \(\sqrt{3}\) m . (ii) (b):   \(\text { In } \Delta A B D, \frac{A B}{B D}=\tan 30^{\circ}\) \(\Rightarrow \frac{25 \sqrt{3}}{B D}=\frac{1}{\sqrt{3}} \Rightarrow B D=75 \mathrm{~m}\) \(\therefore\)  Distance between two positions of car = (75 - 25) m = 50m. (iii) (d): Time taken to cover 50 m distance = 6 sec. \(\therefore\) Time taken to cover 25 m distance = 3 sec. \(\therefore\) Total time taken by car = 6 sec + 3 sec = 9 sec (iv) (c):  \(\text { In } \Delta A B C, \frac{B C}{A C}=\cos 60^{\circ}\) \(\Rightarrow \quad \frac{25}{A C}=\frac{1}{2} \) \(\Rightarrow A C=50 \mathrm{~m}\) (v) (a)

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Case Study on Some Applications of Trigonometry Class 10 Maths PDF

The passage-based questions are commonly known as case study questions. Students looking for Case Study on Some Applications of Trigonometry Class 10 Maths can use this page to download the PDF file. 

The case study questions on Some Applications of Trigonometry are based on the CBSE Class 10 Maths Syllabus, and therefore, referring to the Some Applications of Trigonometry case study questions enable students to gain the appropriate knowledge and prepare better for the Class 10 Maths board examination. Continue reading to know how should students answer it and why it is essential to solve it, etc.

Case Study on Some Applications of Trigonometry Class 10 Maths with Solutions in PDF

Our experts have also kept in mind the challenges students may face while solving the case study on Some Applications of Trigonometry, therefore, they prepared a set of solutions along with the case study questions on Some Applications of Trigonometry.

The case study on Some Applications of Trigonometry Class 10 Maths with solutions in PDF helps students tackle questions that appear confusing or difficult to answer. The answers to the Some Applications of Trigonometry case study questions are very easy to grasp from the PDF - download links are given on this page.

Why Solve Some Applications of Trigonometry Case Study Questions on Class 10 Maths?

There are three major reasons why one should solve Some Applications of Trigonometry case study questions on Class 10 Maths - all those major reasons are discussed below:

• To Prepare for the Board Examination: For many years CBSE board is asking case-based questions to the Class 10 Maths students, therefore, it is important to solve Some Applications of Trigonometry Case study questions as it will help better prepare for the Class 10 board exam preparation.
• Develop Problem-Solving Skills: Class 10 Maths Some Applications of Trigonometry case study questions require students to analyze a given situation, identify the key issues, and apply relevant concepts to find out a solution. This can help CBSE Class 10 students develop their problem-solving skills, which are essential for success in any profession rather than Class 10 board exam preparation.
• Understand Real-Life Applications: Several Some Applications of Trigonometry Class 10 Maths Case Study questions are linked with real-life applications, therefore, solving them enables students to gain the theoretical knowledge of Some Applications of Trigonometry as well as real-life implications of those learnings too.

How to Answer Case Study Questions on Some Applications of Trigonometry?

Students can choose their own way to answer Case Study on Some Applications of Trigonometry Class 10 Maths, however, we believe following these three steps would help a lot in answering Class 10 Maths Some Applications of Trigonometry Case Study questions.

• Read Question Properly: Many make mistakes in the first step which is not reading the questions properly, therefore, it is important to read the question properly and answer questions accordingly.
• Highlight Important Points Discussed in the Clause: While reading the paragraph, highlight the important points discussed as it will help you save your time and answer Some Applications of Trigonometry questions quickly.
• Go Through Each Question One-By-One: Ideally, going through each question gradually is advised so, that a sync between each question and the answer can be maintained. When you are solving Some Applications of Trigonometry Class 10 Maths case study questions make sure you are approaching each question in a step-wise manner.

What to Know to Solve Case Study Questions on Class 10 Some Applications of Trigonometry?

 A few essential things to know to solve Case Study Questions on Class 10 Some Applications of Trigonometry are -

• Basic Formulas of Some Applications of Trigonometry: One of the most important things to know to solve Case Study Questions on Class 10 Some Applications of Trigonometry is to learn about the basic formulas or revise them before solving the case-based questions on Some Applications of Trigonometry.
• To Think Analytically: Analytical thinkers have the ability to detect patterns and that is why it is an essential skill to learn to solve the CBSE Class 10 Maths Some Applications of Trigonometry case study questions.
• Strong Command of Calculations: Another important thing to do is to build a strong command of calculations especially, mental Maths calculations.

Where to Find Case Study on Some Applications of Trigonometry Class 10 Maths?

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• Important Questions for CBSE Class 10 Maths Chapter 8 Introduction to Trigonometry

Class 10 Maths Important Questions Chapter 8 Introduction to Trigonometry: Free PDF Download

CBSE Class 10 Maths Chapter 8 Important Questions revolve around the concept of trigonometric equations at its base. The Class 10 Maths Ch 8 Important Questions by Vedantu come with all the solutions that are drafted in a way, to provide you with a maximum understanding of the subject and figure out the different aspects of trigonometry. The important questions include questions from all the key concepts like basic trigonometry, opposite & adjacent sides in a right-angled triangle, basic trigonometric ratios, and standard values of trigonometric ratios and complementary trigonometric ratios. 

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Study Important Questions for Class 10 Mathematics Chapter 8 - Introduction to Trigonometry

1. If \[x\cos \theta y\sin \theta =a,\,x\sin \theta +y\cos \theta =b\], Prove that \[{{x}^{2}}+{{y}^{2}}={{a}^{2}}+{{b}^{2}}\].

\[x\cos \theta y\sin \theta =a\]        …… (1)

\[x\sin \theta +y\cos \theta =b\]        …… (2)

Squaring and adding the equation (1) and (2) on both sides.

\[{{x}^{2}}{{\cos }^{2}}\theta +{{y}^{2}}{{\sin }^{2}}\theta -2xy\cos \theta \sin \theta +{{x}^{2}}{{\sin }^{2}}\theta +{{y}^{2}}{{\cos }^{2}}\theta +2xy\cos \theta \sin \theta ={{a}^{2}}+{{b}^{2}}\]

\[\Rightarrow {{x}^{2}}\left( {{\cos }^{2}}\theta +{{\sin }^{2}}\theta  \right)+{{y}^{2}}\left( {{\sin }^{2}}\theta +{{\cos }^{2}}\theta  \right)={{a}^{2}}+{{b}^{2}}\]

\[\Rightarrow {{x}^{2}}+{{y}^{2}}={{a}^{2}}+{{b}^{2}}\]

\[\therefore {{x}^{2}}+{{y}^{2}}={{a}^{2}}+{{b}^{2}}\]

Hence proved.

2. Prove that \[{{\sec }^{2}}\theta +\cos e{{c}^{2}}\theta \] Can Never Be Less Than \[2\].

Given: \[{{\sec }^{2}}\theta +\cos e{{c}^{2}}\theta \] 

We know that, ${{\sec }^{2}}\theta =1+{{\tan }^{2}}\theta $ and ${{\operatorname{cosec}}^{2}}\theta =1+{{\cot }^{2}}\theta $.

\[\Rightarrow {{\sec }^{2}}\theta +\cos e{{c}^{2}}\theta =1+{{\tan }^{2}}\theta +1+{{\cot }^{2}}\theta \]

\[\Rightarrow {{\sec }^{2}}\theta +\cos e{{c}^{2}}\theta =2+{{\tan }^{2}}\theta +{{\cot }^{2}}\theta \]

Therefore, \[{{\sec }^{2}}\theta +\cos e{{c}^{2}}\theta \] can never be less than \[2\].

3. If \[\sin \varphi =\dfrac{1}{2}\], show that \[3\cos \varphi -4{{\cos }^{3}}\varphi =0\].

Given: \[\sin \varphi =\dfrac{1}{2}\]

We know that $\sin 30{}^\circ =\dfrac{1}{2}$.

While comparing the angles of $\sin $, we get  

\[\Rightarrow \varphi ={{30}^{\circ }}\]

Substitute \[\varphi ={{30}^{\circ }}\] to get 

\[3\cos \varphi -4{{\cos }^{3}}\varphi =3\cos \left( 30{}^\circ  \right)-4{{\cos }^{3}}\left( 30{}^\circ  \right)\]

\[\Rightarrow 3\left( \dfrac{\sqrt{3}}{2} \right)-4\left( \dfrac{3\sqrt{3}}{8} \right)\Rightarrow 0\]

Therefore, \[3\cos \varphi -4{{\cos }^{3}}\varphi =0\] .

Hence proved .

4. If \[7{{\sin }^{2}}\varphi +3{{\cos }^{2}}\varphi =4\], then Show that \[\tan \varphi =\dfrac{1}{\sqrt{3}}\].

Given: \[7{{\sin }^{2}}\varphi +3{{\cos }^{2}}\varphi =4\]

We know that, ${{\sin }^{2}}\varphi +{{\cos }^{2}}\varphi =1$ and \[\tan \theta =\dfrac{\sin \theta }{\cos \theta }\]

Then, \[7{{\sin }^{2}}\varphi +3{{\cos }^{2}}\varphi =4({{\sin }^{2}}\varphi +{{\cos }^{2}}\varphi )\]

\[\Rightarrow 7{{\sin }^{2}}\varphi -4{{\sin }^{2}}\varphi =4{{\cos }^{2}}\varphi -3{{\cos }^{2}}\varphi \]

\[\Rightarrow 3{{\sin }^{2}}\varphi ={{\cos }^{2}}\varphi \]

\[\Rightarrow \dfrac{{{\sin }^{2}}\varphi }{{{\cos }^{2}}\varphi }=\dfrac{1}{3}\]

\[\Rightarrow {{\tan }^{2}}\varphi =\dfrac{1}{3}\]

\[\Rightarrow \tan \varphi =\dfrac{1}{\sqrt{3}}\]

\[\therefore \tan \varphi =\dfrac{1}{\sqrt{3}}\]

5. If \[\cos \varphi +\sin \varphi =\sqrt{2}\cos \varphi \], Prove that \[\cos \varphi -\sin \varphi =\sqrt{2}\sin \varphi \].

Given : \[\cos \varphi +\sin \varphi =\sqrt{2}\cos \varphi \]

Squaring on both sides, we get

\[\Rightarrow {{\left( \cos \varphi +\sin \varphi  \right)}^{2}}=2{{\cos }^{2}}\varphi \]

\[\Rightarrow {{\cos }^{2}}\varphi +{{\sin }^{2}}\varphi +2\cos \varphi \sin \varphi =2{{\cos }^{2}}\varphi \]

$\Rightarrow {{\sin }^{2}}\varphi =2{{\cos }^{2}}\varphi -{{\cos }^{2}}\varphi -2\cos \varphi \sin \varphi $

$\Rightarrow {{\sin }^{2}}\varphi ={{\cos }^{2}}\varphi -2\cos \varphi \sin \varphi $

Add ${{\sin }^{2}}\varphi $ on both sides

\[\Rightarrow 2{{\sin }^{2}}\varphi ={{\cos }^{2}}\varphi -2\cos \varphi \sin \varphi +{{\sin }^{2}}\varphi \]

\[\Rightarrow 2{{\sin }^{2}}\varphi ={{\left( \cos \varphi -\sin \varphi  \right)}^{2}}\] 

\[\therefore \cos \varphi -\sin \varphi =\sqrt{2}\sin \varphi \]

6. If \[\tan A+\sin A=m\] and \[\tan A-\sin A=n\], then Show that \[{{m}^{2}}-{{n}^{2}}=4\sqrt{mn}\].

Given: 

\[\tan A+\sin A=m\]           …… (1)

\[\tan A-\sin A=n\]            …… (2)

Now to prove \[{{m}^{2}}-{{n}^{2}}=4\sqrt{mn}\].

Take left-hand side

\[{{m}^{2}}-{{n}^{2}}={{\left( \tan A+\sin A \right)}^{2}}-{{\left( \tan A-\sin A \right)}^{2}}\]

\[\Rightarrow {{\tan }^{2}}A+{{\sin }^{2}}A+2\tan A\sin A-{{\tan }^{2}}A-{{\sin }^{2}}A+2\tan A\sin A\]

\[\Rightarrow 4\tan A\sin A\]         

$\therefore {{m}^{2}}-{{n}^{2}}=4\tan A\sin A$      …… (3)

Now take right-hand side

 \[4\sqrt{mn}=4\sqrt{\left( \tan A+\sin A \right)\left( \tan A-\sin A \right)}\]

\[\Rightarrow 4\sqrt{{{\tan }^{2}}A-{{\sin }^{2}}A}\Rightarrow 4\sqrt{\dfrac{{{\sin }^{2}}A}{{{\cos }^{2}}A}-{{\sin }^{2}}A}\]

\[\Rightarrow 4\sqrt{{{\sin }^{2}}A\left( \dfrac{1}{{{\cos }^{2}}A}-1 \right)}\Rightarrow 4\sin A\sqrt{{{\sec }^{2}}A-1}\]

\[\Rightarrow 4\sin A\sqrt{{{\tan }^{2}}A}\Rightarrow 4\sin A\tan A\]

Hence, $4\sqrt{mn}=4\tan A\sin A$

\[\therefore {{m}^{2}}-{{n}^{2}}=4\sqrt{mn}\]

Hence proved. 

7. If \[\sec A=x+\dfrac{1}{4x}\], then prove that \[\sec A+\tan A=2x\] or \[\dfrac{1}{2x}\].

Given: \[\sec A=x+\dfrac{1}{4x}\]

Squaring on both sides.

\[\Rightarrow {{\sec }^{2}}A={{\left( x+\dfrac{1}{4x} \right)}^{2}}\]

We know that, \[{{\sec }^{2}}A=1+{{\tan }^{2}}A\]

\[\Rightarrow 1+{{\tan }^{2}}A={{\left( x+\dfrac{1}{4x} \right)}^{2}}\]

\[\Rightarrow {{\tan }^{2}}A={{\left( x+\dfrac{1}{4x} \right)}^{2}}-1\]

\[\Rightarrow {{\tan }^{2}}A={{x}^{2}}+\dfrac{1}{16{{x}^{2}}}+\dfrac{1}{2}-1\Rightarrow {{x}^{2}}+\dfrac{1}{16{{x}^{2}}}-\dfrac{1}{2}\Rightarrow {{\left( x-\dfrac{1}{4x} \right)}^{2}}\]

Taking square root on both sides,

\[\Rightarrow \tan A=\pm \left( x-\dfrac{1}{4x} \right)\].

Now, find $\sec A+\tan A$

If $\tan A=x-\dfrac{1}{4x}$ means

$\sec A+\tan A=x+\dfrac{1}{4x}+x-\dfrac{1}{4x}\Rightarrow 2x$

$\therefore \sec A+\tan A=2x$

And if $\tan A=-x+\dfrac{1}{4x}$ means

$\sec A+\tan A=x+\dfrac{1}{4x}-x+\dfrac{1}{4x}\Rightarrow \dfrac{2}{4x}\Rightarrow \dfrac{1}{2x}$

$\therefore \sec A+\tan A=\dfrac{1}{2x}$

8. If \[A,B\]are Acute Angles and \[\sin A=\cos B\], then Find the Value of \[A+B\].

Given: $\sin A=\cos B$

We know that $\sin A=\cos \left( 90{}^\circ -A \right)$

While comparing the values to get

$\cos B=\cos \left( 90{}^\circ -A \right)$

$\Rightarrow B=90{}^\circ -A\Rightarrow A+B=90{}^\circ $

\[\therefore A+B={{90}^{\circ }}\].

9. Evaluate the Following Questions:

a. Solve for \[\phi \], if \[\tan 5\phi =1\].

Given: \[\tan 5\phi =1\]

We know that, ${{\tan }^{-1}}\left( 1 \right)=45{}^\circ $

$5\phi ={{\tan }^{-1}}\left( 1 \right)\Rightarrow 45{}^\circ $

$5\phi =45{}^\circ $

$\phi =\dfrac{45{}^\circ }{5}\Rightarrow 9{}^\circ $

\[\because \phi ={{9}^{\circ }}\]

b. Solve for \[\varphi \] , if \[\dfrac{\sin \varphi }{1+\cos \varphi }+\dfrac{1+\cos \varphi }{\sin \varphi }=4\].

Given: \[\dfrac{\sin \varphi }{1+\cos \varphi }+\dfrac{1+\cos \varphi }{\sin \varphi }=4\]

\[\dfrac{{{\sin }^{2}}\varphi +1+{{\cos }^{2}}\varphi +2\cos \varphi }{\sin \varphi \left( 1+\cos \varphi  \right)}=4\]

\[\dfrac{{{\sin }^{2}}\varphi +{{(1+\cos \varphi )}^{2}}}{\sin \varphi (1+\cos \varphi )}=4\]

\[\dfrac{2+2\cos \varphi }{\sin \varphi \left( 1+\cos \varphi  \right)}=4\]

\[\dfrac{2(1+\cos \varphi )}{\sin \varphi (1+\cos \varphi )}=4\]

\[\dfrac{2}{\sin \varphi }=4\]

\[\sin \varphi =\dfrac{1}{2}\]

We know that, $\sin 30{}^\circ =\dfrac{1}{2}$

\[\sin \varphi =\sin {{30}^{\circ }}\Rightarrow \varphi ={{30}^{\circ }}\]

\[\therefore \varphi ={{30}^{\circ }}\].

10. If \[\dfrac{\cos \alpha }{\cos \beta }=m\] and \[\dfrac{\cos \alpha }{\sin \beta }=n\], show that \[({{m}^{2}}+{{n}^{2}}){{\cos }^{2}}\beta ={{n}^{2}}\].

Given: \[\dfrac{\cos \alpha }{\cos \beta }=m\]         …… (1)

\[\dfrac{\cos \alpha }{\sin \beta }=n\]             …… (2)

Squaring equation (1) and (2). We get,

\[{{m}^{2}}=\dfrac{{{\cos }^{2}}\alpha }{{{\cos }^{2}}\beta }\]

\[{{n}^{2}}=\dfrac{{{\cos }^{2}}\alpha }{{{\sin }^{2}}\beta }\]

Now to prove \[({{m}^{2}}+{{n}^{2}}){{\cos }^{2}}\beta ={{n}^{2}}\] ,

Take left-hand side,  

\[({{m}^{2}}+{{n}^{2}}){{\cos }^{2}}\beta =\left( \dfrac{{{\cos }^{2}}\alpha }{{{\cos }^{2}}\beta }+\dfrac{{{\cos }^{2}}\alpha }{{{\sin }^{2}}\beta } \right){{\cos }^{2}}\beta \]

\[=\left( \dfrac{{{\cos }^{2}}\alpha {{\sin }^{2}}\beta +{{\cos }^{2}}\alpha {{\cos }^{2}}\beta }{{{\cos }^{2}}\beta {{\sin }^{2}}\beta } \right){{\cos }^{2}}\beta \]

\[={{\cos }^{2}}\alpha \left( \dfrac{1}{{{\cos }^{2}}\beta {{\sin }^{2}}\beta } \right){{\cos }^{2}}\beta \]

\[=\dfrac{{{\cos }^{2}}\alpha }{{{\sin }^{2}}\beta }\]

\[={{n}^{2}}\]

\[\therefore ({{m}^{2}}+{{n}^{2}}){{\cos }^{2}}\beta ={{n}^{2}}\]

11. If \[7\cos ec\varphi -3\cot \varphi =7\], then prove that \[7\cot \varphi -3\cos ec\varphi =3\].

Given: \[7\cos ec\varphi -3\cot \varphi =7\]

Then prove that, \[7\cot \varphi -3\cos ec\varphi =3\]

\[7\cos ec\varphi -3\cot \varphi =7\]

\[49{{\operatorname{cosec}}^{2}}\varphi +9{{\cot }^{2}}\varphi -42\operatorname{cosec}\varphi \cot \varphi =49\]

We know that, ${{\operatorname{cosec}}^{2}}\varphi =1+{{\cot }^{2}}\varphi $ and ${{\cot }^{2}}\varphi ={{\operatorname{cosec}}^{2}}\varphi -1$.

\[49\left( {{\cot }^{2}}\varphi +1 \right)+9\left( {{\operatorname{cosec}}^{2}}\varphi -1 \right)-42\operatorname{cosec}\varphi \cot \varphi =49\]

\[49{{\cot }^{2}}\varphi +49+9{{\operatorname{cosec}}^{2}}\varphi -9-2\left( 3\operatorname{cosec}\varphi \cdot 7\cot \varphi  \right)=49\]

\[{{\left( 7\cot \varphi -3\operatorname{cosec}\varphi  \right)}^{2}}=49-49+9\]

\[{{\left( 7\cot \varphi -3\operatorname{cosec}\varphi  \right)}^{2}}=9\]

Take square root on both sides, we get

\[\therefore 7\cot \varphi -3\operatorname{cosec}\varphi =3\]

12. Prove that \[2\left( {{\sin }^{6}}\varphi +{{\cos }^{6}}\varphi  \right)\text{ }3\left( {{\sin }^{4}}\varphi +{{\cos }^{4}}\varphi  \right)+1=0\].

Given: \[2\left( {{\sin }^{6}}\varphi +{{\cos }^{6}}\varphi  \right)\text{ }3\left( {{\sin }^{4}}\varphi +{{\cos }^{4}}\varphi  \right)+1=0\]

Let us take left-hand side,

\[\begin{align}&2\left( {{\sin }^{6}}\varphi +{{\cos }^{6}}\varphi  \right)\text{ }3\left( {{\sin }^{4}}\varphi +{{\cos }^{4}}\varphi  \right)+1 \\ & =2\left( {{\left( {{\sin }^{2}}\varphi  \right)}^{3}}+{{\left( {{\cos }^{2}}\varphi  \right)}^{3}} \right)-3\left( {{\left( {{\sin }^{2}}\varphi  \right)}^{2}}+{{\left( {{\cos}^{2}}\varphi  \right)}^{2}} \right)+1 \\ \end{align}\]

\[=2\left[ {{\left( {{\sin }^{2}}\varphi +{{\cos }^{2}}\varphi  \right)}^{3}}-3{{\sin }^{2}}\varphi {{\cos }^{2}}\varphi \left( {{\sin }^{2}}\varphi +{{\cos }^{2}}\varphi  \right) \right]-3\left[ {{\left( {{\sin }^{2}}\varphi +{{\cos }^{2}}\varphi  \right)}^{2}}-2{{\sin }^{2}}\varphi {{\cos }^{2}}\varphi  \right]+1\]

\[=2\left[ 1-3{{\sin }^{2}}\varphi {{\cos }^{2}}\varphi  \right]-3\left[ 1-2{{\sin }^{2}}\varphi {{\cos }^{2}}\varphi  \right]+1\]

\[=2-6{{\sin }^{2}}\varphi {{\cos }^{2}}\varphi -3+6{{\sin }^{2}}\varphi {{\cos }^{2}}\varphi +1\]

Therefore, \[2\left( {{\sin }^{6}}\varphi +{{\cos }^{6}}\varphi  \right)\text{ }3\left( {{\sin }^{4}}\varphi +{{\cos }^{4}}\varphi  \right)+1=0\].

13. If \[\tan \theta =\dfrac{5}{6}\] and \[\theta =\phi ={{90}^{\circ }}\]. What is the value of \[\cot \phi \].

Given: \[\tan \theta =\dfrac{5}{6}\] and \[\theta =\phi ={{90}^{\circ }}\]

We know that, $\tan \theta =\dfrac{1}{\cot \theta }$.

$\cot \phi =\dfrac{1}{\tan \phi }$

$=\dfrac{1}{{5}/{6}\;}$

$=\dfrac{6}{5}$ 

$\therefore \cot \phi =\dfrac{6}{5}$.

14. What is the Value of \[\tan \varphi \] in terms of \[\sin \varphi \] ?

Given: \[\tan \varphi \]

We know that, \[\tan \varphi =\dfrac{\sin \varphi }{\cos \varphi }\] and \[{{\cos }^{2}}\varphi +{{\sin }^{2}}\varphi =1\]

\[\tan \varphi =\dfrac{\sin \varphi }{\cos \varphi }\]

\[\therefore \tan \varphi =\dfrac{\sin \varphi }{\sqrt{1-{{\sin }^{2}}\varphi }}\]

15. If \[\sec \varphi +\tan \varphi =4\], Find the Value of \[\sin \varphi \], \[\cos \varphi \].

Given: \[\sec \varphi +\tan \varphi =4\]

\[\dfrac{1}{\cos \varphi }+\dfrac{\sin \varphi }{\cos \varphi }=4\]

\[\dfrac{1+\sin \varphi }{\cos \varphi }=4\]

$1+\sin \varphi =4\cos \varphi $

${{\left( 1+\sin \varphi  \right)}^{2}}={{\left( 4\cos \varphi  \right)}^{2}}$

$1+2\sin \varphi +{{\sin }^{2}}\varphi =16{{\cos }^{2}}\varphi $

$1+2\sin \varphi +{{\sin }^{2}}\varphi =16\left( 1-{{\sin }^{2}}\varphi  \right)$

$1+2\sin \varphi +{{\sin }^{2}}\varphi =16-16{{\sin }^{2}}\varphi $

$17{{\sin }^{2}}\varphi +2\sin \varphi -15=0$

$17{{\sin }^{2}}\varphi +17\sin \varphi -15\sin \varphi -15=0$

$17\sin \varphi \left( \sin \varphi +1 \right)-15\left( \sin \varphi +1 \right)=0$

$\left( \sin \varphi +1 \right)\left( 17\sin \varphi -15 \right)=0$

If $\sin \varphi +1=0$

Hence, $\sin \varphi =-1$ is not possible.

Then, $17\sin \varphi -15=0$

$\therefore \sin \varphi =\dfrac{15}{17}$

Now find $\cos \varphi $

Substitute the value of $\sin \varphi $

\[1+\dfrac{15}{17}=4\cos \varphi \]

\[\dfrac{32}{17}=4\cos \varphi \]

\[\Rightarrow \cos \varphi =\dfrac{32}{17\left( 4 \right)}\Rightarrow \dfrac{8}{17}\]

\[\therefore \cos \varphi =\dfrac{8}{17}\]

Short Answer Questions (2 Marks)

1. In \[\Delta ABC\] , Right Angled at \[B,AB=24cm,BC=7cm\].

Determine the Following Equations:    

(i) \[\sin A,\cos A\]

Let us draw a right-angled triangle \[ABC\], right angled at \[B\].

Using Pythagoras theorem, find $AC$.

\[A{{C}^{2}}=A{{B}^{2}}+B{{C}^{2}}\]

\[={{(24)}^{2}}+{{(7)}^{2}}\]

\[=576+49\]

\[\therefore AC=25cm\]

\[\sin A=\dfrac{BC}{AC}=\dfrac{7}{25}\]

\[\therefore \sin A=\dfrac{7}{25}\]

 \[\cos A=\dfrac{AB}{AC}=\dfrac{24}{25}\]

\[\therefore \cos A=\dfrac{24}{25}\]

(ii) \[\sin C,\cos C\]

\[\sin C=\dfrac{AB}{AC}=\dfrac{24}{25}\]       

\[\therefore \sin C=\dfrac{24}{25}\]

\[\cos C=\dfrac{BC}{AC}=\dfrac{7}{25}\]

\[\therefore \cos C=\dfrac{7}{25}\]

2. In Adjoining Figure, Find the Value of \[\tan P-\cot R\].     

Using Pythagoras theorem,

$P{{R}^{2}}=P{{Q}^{2}}+Q{{R}^{2}}$

\[{{(13)}^{2}}={{(12)}^{2}}+Q{{R}^{2}}\]

\[Q{{R}^{2}}=169-144\Rightarrow 25\]

\[\therefore QR=5cm\]

Then find \[\tan P-\cot R\] ,

First find the value of $\tan P$

$\tan P=\dfrac{side\,opp.\,to\,\angle P}{side\,adj.\,to\,\angle P}=\dfrac{QR}{PQ}=\dfrac{5}{12}$

$\therefore \tan P=\dfrac{5}{12}$

Now find the value of $\cot R$

We know that, $\tan R=\dfrac{1}{\cot R}$

For that we need to first find the value of $\tan R$

$\tan R=\dfrac{side\,opp.\,to\,\angle R}{side\,adj.\,to\,\angle R}=\dfrac{PQ}{QR}=\dfrac{12}{5}$

$\therefore \cot R=\dfrac{5}{12}$

Then, 

\[\tan P-\cot R=\dfrac{QR}{PQ}-\dfrac{QR}{PQ}\]

\[\Rightarrow \dfrac{5}{12}-\dfrac{5}{12}\Rightarrow 0\]

\[\therefore \tan P-\cot R=0\].

3. If \[\sin A=\dfrac{3}{4}\], Calculate the Value of \[\cos A\] and \[\tan A\].                                                         

Given that the triangle \[ABC\] in which \[\angle B={{90}^{\circ }}\]

Let us take \[BC=3k\] and \[AC=4k\]

Then using Pythagoras theorem,

\[AB=\sqrt{{{(AC)}^{2}}-{{(BC)}^{2}}}\]

 \[\Rightarrow \sqrt{{{(4k)}^{2}}-{{(3k)}^{2}}}\]

 \[\Rightarrow \sqrt{16k-9k}\]

\[\Rightarrow k\sqrt{7}\]

$\therefore AB=k\sqrt{7}$

Calculate the value of \[\cos A\]  

$\cos A=\dfrac{AB}{AC}=\dfrac{k\sqrt{7}}{4k}=\dfrac{\sqrt{7}}{4}$

$\therefore \cos A=\dfrac{\sqrt{7}}{4}$

And calculate the value of \[\tan A\]

$\tan A=\dfrac{BC}{AB}=\dfrac{3k}{k\sqrt{7}}=\dfrac{3}{\sqrt{7}}$

$\therefore \tan A=\dfrac{3}{\sqrt{7}}$

4. Given \[15\cot A=8\], Find the Values of \[\sin A\] and \[\sec A\].                                                           

Given: \[15\cot A=8\]

Let us assume a triangle \[ABC\] in which \[\angle B={{90}^{\circ }}\]. 

\[15\cot A=8\]

\[\Rightarrow \cot A=\dfrac{8}{15}\]

Since $\cot A=\dfrac{adj}{hyp}=\dfrac{AB}{BC}$.

Let us draw the triangle.

Now, \[AB=8k\] and \[BC=15k\].

Using Pythagoras theorem, find the value of $AC$.

\[AC=\sqrt{{{(AB)}^{2}}+{{(BC)}^{2}}}\]

\[\Rightarrow \sqrt{{{(8k)}^{2}}+{{(15k)}^{2}}}\]

$\Rightarrow \sqrt{64{{k}^{2}}+225{{k}^{2}}}$

$\Rightarrow \sqrt{289{{k}^{2}}}$

\[\Rightarrow 17k\]

$\therefore AC=17k$

Now, find the values of \[\sin A\] and \[\sec A\] .

\[\sin A=\dfrac{BC}{AC}=\dfrac{15k}{17k}=\dfrac{15}{17}\]

$\therefore \sin A=\dfrac{15}{17}$

\[\sec A=\dfrac{AC}{AB}=\dfrac{17k}{8k}=\dfrac{17}{8}\]

 \[\therefore \sec A=\dfrac{17}{8}\]

5. If \[\angle A\] and \[\angle \,B\] are Acute Angles Such That \[\cos A=\cos B\], then show that \[\angle A=\angle \,B\]

Given: \[\cos A=\cos B\]

In right triangle \[ABC\],

\[\cos A=\dfrac{side\,adj.\,A}{hyp.}=\dfrac{AC}{AB}\]  …… (1)

And, \[\cos B=\dfrac{side\,adj.\,B}{hyp.}=\dfrac{BC}{AB}\] …… (2)

Then, \[\cos A=\cos B\]

Now, equate equation (1) and (2).

\[\Rightarrow \dfrac{AC}{AB}=\dfrac{BC}{AB}\]

$\Rightarrow AC=BC$

\[\Rightarrow \angle A=\angle B\].

Therefore, Angles opposite to equal sides are equal.

6. State Whether the Following are True or False. Justify Your Answer.           

 The Value of \[\tan A\] is Always Less than \[1\].

False because sides of a right triangle may have any length, so \[\tan A\] may have any value. For example, $\tan A=\dfrac{BC}{AB}=\dfrac{15}{10}=\dfrac{3}{2}=1.5$. 

  \[\sec A=\dfrac{12}{5}\] for Some Value of Angle \[A\] .

True as \[\sec A\] is always greater than \[1\]. For example, $\sec A=\dfrac{hyp.}{side\,adj.\,A}$ . As hypotenuse will be the largest side. So, it is true.

  \[\cos A\] is the Abbreviation Used for the Cosecant of Angle \[A\] .

False as \[\cos A\] is the abbreviation of \[\operatorname{cosineA}\]. Because $\cos A$ means cosine of angle $A$ and $\operatorname{co}\sec A$ means cosecant of angle $A$.

 \[\cot A\] is the Product of \[\cot \]and \[A\].

False as \[\cot A\] is not the product of \[cot\] and \[A\]. \[cot\] without \[A\] doesn’t have meaning. 

 \[\sin \theta =\dfrac{4}{3}\] for Some Angle \[\theta \].

Ans:  

False as \[\sin \theta \] cannot be greater than \[1\]. For example, $\sin \theta =\dfrac{side\,opp.\,\theta }{hyp}$. Since the hypotenuse is the largest side. So, \[\sin \theta \] will be less than $1$.

7. Evaluate the Following Equations: 

i. \[\dfrac{\sin {{18}^{\circ }}}{\cos {{72}^{\circ }}}\]

Given: \[\dfrac{\sin {{18}^{\circ }}}{\cos {{72}^{\circ }}}\]

We know that, $\sin \left( 90{}^\circ -\theta  \right)=\cos \theta $

\[\dfrac{\sin {{18}^{\circ }}}{\cos {{72}^{\circ }}}=\dfrac{\sin ({{90}^{\circ }}-{{72}^{\circ }})}{\cos {{72}^{\circ }}}=\dfrac{\cos {{72}^{\circ }}}{\cos {{72}^{\circ }}}=1\]                   

$\therefore \dfrac{\sin 18{}^\circ }{\cos 72{}^\circ }=1$

ii. \[\dfrac{\tan {{26}^{\circ }}}{\cot {{64}^{\circ }}}\]

Given: \[\dfrac{\tan {{26}^{\circ }}}{\cot {{64}^{\circ }}}\]

We know that, $\tan \left( 90{}^\circ -\theta  \right)=\cot \theta $

\[\dfrac{\tan {{26}^{\circ }}}{\cot {{64}^{\circ }}}=\dfrac{\tan ({{90}^{\circ }}-{{64}^{\circ }})}{\cot {{64}^{\circ }}}=\dfrac{\cot {{64}^{\circ }}}{\cot {{64}^{\circ }}}=1\]

$\therefore \dfrac{\tan 26{}^\circ }{\cot 64{}^\circ }=1$

iii. \[\cos {{48}^{\circ }}-\sin {{42}^{\circ }}\]

Given: \[\cos {{48}^{\circ }}-\sin {{42}^{\circ }}\]

We know that, $\cos \left( 90{}^\circ -\theta  \right)=\sin \theta $

\[\Rightarrow \cos ({{90}^{\circ }}-{{42}^{\circ }})-\sin {{42}^{\circ }}\]

\[\Rightarrow \sin {{42}^{\circ }}-\sin {{42}^{\circ }}\Rightarrow 0\]

 \[\therefore \cos 48{}^\circ -\sin 42{}^\circ =0\]

iv. \[\operatorname{cosec}31{}^\circ -\sec 59{}^\circ \]

Given: \[\operatorname{cosec}31{}^\circ -\sec 59{}^\circ \]

We know that, $\operatorname{cosec}\left( 90{}^\circ -\theta  \right)=\sec \theta $

\[\Rightarrow \operatorname{cosec}(({{90}^{\circ }}-{{59}^{\circ }})-\sec {{59}^{\circ }}\]

\[\Rightarrow \sec {{59}^{\circ }}-\sec {{59}^{\circ }}\Rightarrow 0\]

\[\therefore \operatorname{cosec}31{}^\circ -\sec 59{}^\circ =0\]

1. Show that the Following Equations:                       

\[\tan {{48}^{\circ }}\tan {{23}^{\circ }}\tan {{42}^{\circ }}\tan {{67}^{\circ }}=1\]

Given: \[\tan {{48}^{\circ }}\tan {{23}^{\circ }}\tan {{42}^{\circ }}\tan {{67}^{\circ }}=1\]

We know that, $\tan \left( 90{}^\circ -\theta  \right)=\cot \theta $.

Now let us take left-hand side,

\[\tan {{48}^{\circ }}\tan {{23}^{\circ }}\tan {{42}^{\circ }}\tan {{67}^{\circ }}\]

\[\Rightarrow \tan ({{90}^{\circ }}-{{42}^{\circ }})\tan ({{90}^{\circ }}-{{67}^{\circ }})\tan {{42}^{\circ }}\tan {{67}^{\circ }}\]

\[\Rightarrow \cot {{42}^{\circ }}\cot {{67}^{\circ }}\tan {{42}^{\circ }}\tan {{67}^{\circ }}\]

\[\Rightarrow \dfrac{1}{\tan {{42}^{\circ }}}.\dfrac{1}{\tan {{67}^{\circ }}}.\tan {{42}^{\circ }}\tan {{67}^{\circ }}\]

\[\Rightarrow 1\] is equal to R.H.S

\[\therefore \tan {{48}^{\circ }}\tan {{23}^{\circ }}\tan {{42}^{\circ }}\tan {{67}^{\circ }}=1\]

(ii) \[\cos {{38}^{\circ }}\cos {{52}^{\circ }}-\sin {{38}^{\circ }}\sin {{52}^{\circ }}=0\]

Given: \[\cos {{38}^{\circ }}\cos {{52}^{\circ }}-\sin {{38}^{\circ }}\sin {{52}^{\circ }}=0\]

Now let us take left-hand side, 

\[\cos {{38}^{\circ }}cos{{52}^{\circ }}-\sin {{38}^{\circ }}\sin {{52}^{\circ }}\]

\[\Rightarrow \cos ({{90}^{\circ }}-{{52}^{\circ }})\cos ({{90}^{\circ }}-{{38}^{\circ }})-\sin {{38}^{\circ }}\sin {{52}^{\circ }}\]

\[\Rightarrow \sin {{52}^{\circ }}\sin {{38}^{\circ }}-\sin {{38}^{\circ }}\sin {{52}^{\circ }}\]

\[\Rightarrow 0\] is equal to R.H.S

$\therefore \cos {{38}^{\circ }}cos{{52}^{\circ }}-\sin {{38}^{\circ }}\sin {{52}^{\circ }}=0$

2. If  \[\tan 2A=\cot (A-{{18}^{\circ }})\] where \[2A\] is an Acute Angle, Find the Value of \[A\]. 

Given: \[\tan 2A=\cot (A-{{18}^{\circ }})\]

We know that, $\cot \left( 90{}^\circ -\theta  \right)=\tan \theta $

\[\Rightarrow \cot ({{90}^{\circ }}-2A)=\cot (A-{{18}^{\circ }})\]

Now equalise the angles,

\[{{90}^{\circ }}-2A=A-{{18}^{\circ }}\]

\[-2A-A=-{{18}^{\circ }}-{{90}^{\circ }}\]

\[-3A=-{{108}^{\circ }}\]

\[A=\dfrac{108{}^\circ }{3}\]

$\therefore A=36{}^\circ $

3. If \[\tan A=\cot B\], then Prove That \[A+B={{90}^{\circ }}\].                                                            

Given: \[\tan A=\cot B\]

\[\cot ({{90}^{\circ }}-A)=\cot B\]

\[{{90}^{\circ }}-A=B\]

\[\Rightarrow A+B={{90}^{\circ }}\]

$\therefore A+B=90{}^\circ $

4. If \[\sec 4A=\operatorname{cosec}(A-{{20}^{\circ }})\], Where \[4A\] is an Acute Angle, Then Find the Value of $A$.

Ans:  

Given: \[\sec 4A=\operatorname{cosec}(A-{{20}^{\circ }})\]

\[\Rightarrow \operatorname{cosec}({{90}^{\circ }}-4A)=\operatorname{cosec}(A-{{20}^{\circ }})\]

\[{{90}^{\circ }}-4A=A-{{20}^{\circ }}\]

\[-4A-A=-{{20}^{\circ }}-{{90}^{\circ }}\]

\[-5A=-{{110}^{\circ }}\]

\[A=\dfrac{{{110}^{\circ }}}{5}\]

\[\therefore A={{22}^{\circ }}\]

5. If \[A,B\] and \[C\] are Interior Angles of a \[\Delta ABC\], then Show That \[\sin \left( \dfrac{B+C}{2} \right)=\cos \dfrac{A}{2}\].

Given: \[A,B\] and \[C\] are interior angles of a \[\Delta ABC\].

We know that, \[A+B+C={{180}^{\circ }}\].

Let us consider, 

\[\dfrac{A+B+C}{2}={{90}^{\circ }}\]

\[\Rightarrow \dfrac{B+C}{2}={{90}^{\circ }}-\dfrac{A}{2}\]

Multiply $\sin $ on both sides,

\[\sin \left( \dfrac{B+C}{2} \right)=\sin \left( {{90}^{\circ }}-\dfrac{A}{2} \right)\]

\[\therefore \sin \left( \dfrac{B+C}{2} \right)=\cos \dfrac{A}{2}\]

6. Express \[\sin {{67}^{\circ }}+\cos {{75}^{\circ }}\] in terms of trigonometric ratios of angles between  \[{{0}^{\circ }}\] and \[{{45}^{\circ }}\].                                                   

Given : \[\sin {{67}^{\circ }}+\cos {{75}^{\circ }}\].

We know that, $\sin \left( 90{}^\circ -\theta  \right)=\cos \theta $ and $\cos \left( 90{}^\circ -\theta  \right)=\sin \theta $.

\[\sin {{67}^{\circ }}+\cos {{75}^{\circ }}=\sin ({{90}^{\circ }}-{{23}^{\circ }})+\cos ({{90}^{\circ }}-{{15}^{\circ }})\]

\[=\cos {{23}^{\circ }}+\sin {{15}^{\circ }}\]

\[\therefore \cos {{23}^{\circ }}+\sin {{15}^{\circ }}\] is the required value.

7. Express the Trigonometric Ratios \[\sin A,\sec A\] and \[\tan A\] in Terms of \[\cot A\]. 

Find the value of \[\sin A\] in terms of \[\cot A\].

By using identity \[{{\operatorname{cosec}}^{2}}A-{{\cot }^{2}}A=1\].

Then, use $\operatorname{cosec}A=\dfrac{1}{\sin A}$

\[\Rightarrow {{\operatorname{cosec}}^{2}}A=1+{{\cot }^{2}}A\]

\[\Rightarrow \dfrac{1}{{{\sin }^{2}}A}=1+{{\cot }^{2}}A\]

\[\Rightarrow {{\sin }^{2}}A=\dfrac{1}{1+{{\cot }^{2}}A}\]

\[\Rightarrow \sin A=\dfrac{1}{\sqrt{1+{{\cot }^{2}}A}}\]

\[\therefore \sin A=\dfrac{1}{\sqrt{1+{{\cot }^{2}}A}}\]

Now find the value for \[\sec A\] in terms of \[\cot A\].

Using identity \[{{\sec }^{2}}A-{{\tan }^{2}}A=1\]

\[\Rightarrow {{\sec }^{2}}A=1+{{\tan }^{2}}A\]

\[\Rightarrow {{\sec }^{2}}A=1+\dfrac{1}{{{\cot }^{2}}A}\]

\[\Rightarrow {{\sec }^{2}}A=\dfrac{{{\cot }^{2}}A+1}{{{\cot }^{2}}A}\]

\[\Rightarrow \sec A=\dfrac{\sqrt{1+{{\cot }^{2}}A}}{\cot A}\] 

\[\therefore \sec A=\dfrac{\sqrt{1+{{\cot }^{2}}A}}{\cot A}\]

And find the value for \[\tan A\] in terms of \[\cot A\]

By trigonometric ratio property, $\tan A=\dfrac{1}{\cot A}$

Hence, \[\tan A=\dfrac{1}{\cot A}\]

Therefore, \[\sin A,\sec A\] and \[\tan A\] are founded in terms of \[\cot A\].  

8. Write the Other Trigonometric Ratios of $A$ in Terms of \[\sec A\] .                        

Find the value of \[\sin A\] in terms of \[\sec A\]

By using identity,\[{{\sin }^{2}}A+{{\cos }^{2}}A=1\]

\[\Rightarrow {{\sin }^{2}}A=1-{{\cos }^{2}}A\]

\[\Rightarrow {{\sin }^{2}}A=1-\dfrac{1}{{{\sec }^{2}}A}\]

\[\Rightarrow {{\sin }^{2}}A=\dfrac{{{\sec }^{2}}A-1}{{{\sec }^{2}}A}\]

\[\Rightarrow \sin A=\dfrac{\sqrt{{{\sec }^{2}}A-1}}{\sec A}\]

\[\therefore \sin A=\dfrac{\sqrt{{{\sec }^{2}}A-1}}{\sec A}\]

Now find the value for \[\cos A\] in terms of \[\sec A\],

By trigonometric ratio property, $\cos A=\dfrac{1}{\sec A}$

$\therefore \cos A=\dfrac{1}{\sec A}$

Find the value for $\tan A$ in terms of \[\sec A\],

By using identity, \[{{\sec }^{2}}A-{{\tan }^{2}}A=1\]

\[\Rightarrow {{\tan }^{2}}A={{\sec }^{2}}A-1\]

\[\Rightarrow \tan A=\sqrt{{{\sec }^{2}}A-1}\]

\[\therefore \tan A=\sqrt{{{\sec }^{2}}A-1}\]

Find the value for \[\operatorname{cosec}A\] in terms of \[\sec A\]

By trigonometric ratio property, \[\operatorname{cosec}A=\dfrac{1}{\sin A}\]

\[\Rightarrow \operatorname{cosec}A=\dfrac{1}{\sin A}\]

Substitute the value of \[\sin A=\dfrac{\sqrt{{{\sec }^{2}}A-1}}{\sec A}\].

\[\Rightarrow \operatorname{cosec}A=\dfrac{1}{\dfrac{\sqrt{{{\sec }^{2}}A-1}}{\sec A}}\]

\[\Rightarrow \operatorname{cosec}A=\dfrac{\sec A}{\sqrt{{{\sec }^{2}}A-1}}\]

\[\therefore \operatorname{cosec}A=\dfrac{\sec A}{\sqrt{{{\sec }^{2}}A-1}}\]

Finally, find the value for \[\cot A\] in terms of \[\sec A\]

By trigonometric ratio property, \[\cot A=\dfrac{1}{\tan A}\]

$\Rightarrow \cot A=\dfrac{1}{\tan A}$

Substitute the value of \[\tan A=\sqrt{{{\sec }^{2}}A-1}\]

\[\Rightarrow \cot A=\dfrac{1}{\sqrt{{{\sec }^{2}}A-1}}\]

$\therefore \cot A=\dfrac{1}{\sqrt{{{\sec }^{2}}A-1}}$.

9. Evaluate the Following Equations:                                    

(i) \[\dfrac{{{\sin }^{2}}{{63}^{\circ }}+{{\sin }^{2}}{{27}^{\circ }}}{{{\cos }^{2}}{{17}^{\circ }}+{{\cos }^{2}}{{73}^{\circ }}}\]

Given: \[\dfrac{{{\sin }^{2}}{{63}^{\circ }}+{{\sin }^{2}}{{27}^{\circ }}}{{{\cos }^{2}}{{17}^{\circ }}+{{\cos }^{2}}{{73}^{\circ }}}\]

We know that, \[\sin ({{90}^{\circ }}-\theta )=\cos \theta ,\cos ({{90}^{\circ }}-\theta )=\sin \theta \] and  \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\]

\[\Rightarrow \dfrac{{{\sin }^{2}}{{63}^{\circ }}+{{\sin }^{2}}({{90}^{\circ }}-{{63}^{\circ }})}{{{\cos }^{2}}({{90}^{\circ }}-{{73}^{\circ }})+{{\cos }^{2}}{{73}^{\circ }}}\]

\[\Rightarrow \dfrac{{{\sin }^{2}}{{63}^{\circ }}+{{\cos }^{2}}{{63}^{\circ }}}{{{\sin }^{2}}{{73}^{\circ }}+{{\cos }^{2}}{{73}^{\circ }}}\]         

\[\Rightarrow \dfrac{1}{1}\Rightarrow 1\]                      

$\therefore \dfrac{{{\sin }^{2}}{{63}^{\circ }}+{{\sin }^{2}}{{27}^{\circ }}}{{{\cos }^{2}}{{17}^{\circ }}+{{\cos }^{2}}{{73}^{\circ }}}=1$         

(ii) \[\sin {{25}^{\circ }}\cos {{65}^{\circ }}+\cos {{25}^{\circ }}\sin {{65}^{\circ }}\]

Given: \[\sin {{25}^{\circ }}\cos {{65}^{\circ }}+\cos {{25}^{\circ }}\sin {{65}^{\circ }}\]

We know that, $\sin \left( A+B \right)=\sin A\cos B+\cos A\sin B$

\[\Rightarrow \sin \left( 25{}^\circ +65{}^\circ  \right)\]

\[\Rightarrow \sin 90{}^\circ \]                    

\[\Rightarrow 1\]

$\therefore \sin {{25}^{\circ }}\cos {{65}^{\circ }}+\cos {{25}^{\circ }}\sin {{65}^{\circ }}=1$

10. Show that Any Positive Odd Integer Is of the Form \[\mathbf{6q}\text{ }+\text{ }\mathbf{1}\], or \[\mathbf{6q}\text{ }+\text{ }\mathbf{3},\]or \[\mathbf{6q}\text{ }+\text{ }\mathbf{5}\], where \[q\] is some integer.                                                      

Let \[a\] be any positive integer and \[b=\text{ }6\]. 

Then, by Euclid’s algorithm,

\[a=6q+r\] for some integer \[q\ge 0\], and \[r=0,1,2,3,4,5\] because \[0\le r<\text{ }6\].

Therefore, \[a=6q\] or \[6q+\text{ }1\] or \[6q+\text{ }2\]or \[6q\text{ }+3\]or \[6q+\text{ }4\]or \[6q+\text{ }5\]

Also, \[6q+1=2\times 3q+1=2{{k}_{1}}+1,\] where \[{{k}_{1}}\] is a positive integer

\[6q+3=(6q+2)+1=2(3q+1)+1=2{{k}_{2}}+\text{ }1,\]Where \[{{k}_{2}}\]is an integer

\[6q+5=(6q+4)+1=2(3q+2)+1=2{{k}_{3}}+1\], where \[{{k}_{3}}\] is an integer

Clearly, \[6q+1,6q+3,6q+5\] are of the form \[2k+\text{ }1,\]where \[k\]an integer is.

Therefore, \[6q+1,6q+3,6q+5\] are not exactly divisible by \[2\].

Hence, these expressions of numbers are odd numbers.

And therefore, any odd integer can be expressed in the form \[6q+1\], or \[6q+3\], or\[6q+5.\]

11. An Army Contingent of \[\mathbf{616}\] Members are to March Behind an Army Band of \[\mathbf{32}\] Members in a Parade. The Two Groups Are to March in the Same Number of Columns. What Is the Maximum Number of Columns in Which They Can March?

We have to find the \[HCF\left( 616,\text{ }32 \right)\] to find the maximum number of columns in which they can march. 

To find the HCF, we can use Euclid’s algorithm.

\[616=32\times 19+8\]

\[\Rightarrow 32=8\times 4+0\]

Hence,  \[HCF\left( 616,\text{ }32 \right)\] is \[8\].

Therefore, they can march in \[8\] columns each.

12. Use Euclid’s Division Lemma to Show That the Square of Any Positive Integer Is Either of Form \[\mathbf{3}m\]or \[\mathbf{3}m+\mathbf{1}\] for some integer \[m\]. 

[Hint: Let \[x\] be any positive integer then it is of the form\[\mathbf{3}q,\mathbf{3}q+\mathbf{1}\] or \[\mathbf{3}q+\mathbf{2}\text{ }\]. Now square each of these and show that they can be rewritten in the form \[\mathbf{3}m\]or \[\mathbf{3}m+\mathbf{1}\].]

Let\[\text{ }a\] be any positive integer and \[b=3\].

Then \[a=3q+r\] for some integer \[q\ge 0\]

And \[r=0,1,2\] because \[0\le r<3\]

Therefore, \[a=3q\] or \[3q+1\] or \[3q+2\]

\[\Rightarrow {{a}^{2}}={{(3q)}^{2}}\] or \[{{(3q+1)}^{2}}\] or \[{{(3q+2)}^{2}}\]

\[\Rightarrow {{a}^{2}}={{(9q)}^{2}}\] or \[9{{q}^{2}}+6q+1\] or \[9{{q}^{2}}+12q+4\]

\[\Rightarrow {{a}^{2}}=3\times {{(3q)}^{2}}\] or \[3(3{{q}^{2}}+2q)+1\] or \[3(3{{q}^{2}}+4q+1)+1\]

\[\Rightarrow a=3{{k}_{1}}\] or \[3{{k}_{2}}+1\] or \[3{{k}_{3}}+1\]

Where \[{{k}_{1}},{{k}_{2}},{{k}_{3}}\] are some positive integers 

Hence, it can be said that the square of any positive integer is either of the form \[3m\]or \[3m+1\].

Short Answer Questions (3 Marks)

1. Given \[\sec \theta =\dfrac{13}{12}\] , Calculate the Values for All Other Trigonometric Ratios. 

Given:   \[\sec \theta =\dfrac{13}{12}\]

Let us consider a triangle \[ABC\] in which \[\angle A=\theta \] and \[\angle B={{90}^{\circ }}\]

Let \[AB=12k\] and \[AC=13k\]

Then, find the value of $BC$

\[BC=\sqrt{{{(AC)}^{2}}-{{(AB)}^{2}}}\]

\[\begin{align}&\Rightarrow\sqrt{{{(13k)}^{2}}-{{(12k)}^{2}}}\\&\Rightarrow \sqrt{69{{k}^{2}}-144{{k}^{2}}} \\ & \Rightarrow \sqrt{25{{k}^{2}}} \\ & \Rightarrow 5k \\ \end{align}\]

\[\therefore BC=5k\]

Since, \[\sec \theta =\dfrac{13}{12}\]

Similarly, 

\[\sin \theta =\dfrac{BC}{AC}=\dfrac{5k}{13k}=\dfrac{5}{13}\]

\[\cos \theta =\dfrac{AB}{AC}=\dfrac{12k}{13k}=\dfrac{12}{13}\]

\[\tan \theta =\dfrac{BC}{AB}=\dfrac{5k}{12k}=\dfrac{5}{12}\]

\[\cot \theta =\dfrac{AB}{BC}=\dfrac{12k}{5k}=\dfrac{12}{5}\]

\[\cos ec\theta =\dfrac{AC}{BC}=\dfrac{13k}{5k}=\dfrac{13}{5}\]

2. If \[\cot \theta =\dfrac{7}{8}\] , then Evaluate the Followings Equations :                                                                                        

i. \[\dfrac{(1+\sin \theta )(1-\sin \theta )}{(1+\cos \theta )(1-\cos \theta )}\]

Given: \[\cot \theta =\dfrac{7}{8}\]

Then, \[AB=7k\] and \[BC=8k\]

Using Pythagoras theorem, find $AC$

\[AC=\sqrt{{{(BC)}^{2}}+{{(AB)}^{2}}}\]

\[\Rightarrow \sqrt{{{(8k)}^{2}}+{{(7k)}^{2}}}\]

\[\Rightarrow \sqrt{64{{k}^{2}}+49{{k}^{2}}}\]

\[\Rightarrow \sqrt{113{{k}^{2}}}\]

\[\Rightarrow \sqrt{113}k\]

\[\therefore AC=\sqrt{113}k\]

Now find the value of trigonometric ratios.

\[\sin \theta =\dfrac{BC}{AC}=\dfrac{8k}{\sqrt{113}k}=\dfrac{8}{\sqrt{113}}\] and \[\cos \theta =\dfrac{AB}{AC}=\dfrac{7k}{\sqrt{113}k}=\dfrac{7}{\sqrt{113}}\].

\[\dfrac{(1+\sin \theta )(1-\sin \theta )}{(1+\cos \theta )(1-\cos \theta )}\]

We know that, ${{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1$

\[\Rightarrow \dfrac{1-{{\sin }^{2}}\theta }{1-{{\cos }^{2}}\theta }\Rightarrow \dfrac{{{\cos }^{2}}\theta }{{{\sin }^{2}}\theta }\]

\[\Rightarrow \dfrac{\dfrac{49}{113}}{\dfrac{64}{113}}\Rightarrow \dfrac{49}{113}\cdot \dfrac{113}{64}\Rightarrow \dfrac{49}{64}\]

$\therefore \dfrac{(1+\sin \theta )(1-\sin \theta )}{(1+\cos \theta )(1-\cos \theta )}=\dfrac{49}{64}$

ii. \[{{\cot }^{2}}\theta \]

Given: \[{{\cot }^{2}}\theta \]

We know that, $\cot \theta =\dfrac{\cos \theta }{\sin \theta }$

\[\Rightarrow \dfrac{{{\cos }^{2}}\theta }{{{\sin }^{2}}\theta }\Rightarrow \dfrac{\dfrac{49}{113}}{\dfrac{64}{113}}\Rightarrow \dfrac{49}{64}\]

$\therefore {{\cot }^{2}}\theta =\dfrac{49}{64}$

Hence, $\dfrac{\left( 1+\sin \theta  \right)\left( 1-\sin \theta  \right)}{\left( 1+\cos \theta  \right)\left( 1-\cos \theta  \right)}$ and ${{\cot }^{2}}\theta $ are same.

3. If \[3\cot A=4\], then show that  \[\dfrac{1-{{\tan }^{2}}A}{1+{{\tan }^{2}}A}={{\cos }^{2}}A-{{\sin }^{2}}A\].              

Given: \[3\cot A=4\]

Let us consider a triangle \[ABC\] in which  \[\angle B={{90}^{\circ }}\]

Then, \[3\cot A=4\] 

\[\Rightarrow \cot A=\dfrac{4}{3}\]

Let \[AB=4k\] and \[BC=3k\]

 \[\Rightarrow \sqrt{{{(3k)}^{2}}+{{(4k)}^{2}}}=\sqrt{16{{k}^{2}}+9{{k}^{2}}}\]

\[\Rightarrow \sqrt{25{{k}^{2}}}=5k\]

\[\therefore AC=5k\]

\[\sin A=\dfrac{BC}{AC}=\dfrac{3k}{5k}=\dfrac{3}{5}\]  , \[\cos A=\dfrac{AB}{AC}=\dfrac{4k}{5k}=\dfrac{4}{5}\] and \[\tan A=\dfrac{BC}{AB}=\dfrac{3k}{4k}=\dfrac{3}{4}\].

To prove: \[\dfrac{1-{{\tan }^{2}}A}{1+{{\tan }^{2}}A}={{\cos }^{2}}A-{{\sin }^{2}}A\]

Let us take left-hand side 

L.H.S \[=\dfrac{1-{{\tan }^{2}}A}{1+{{\tan }^{2}}A}\]

Substitute the value of $\tan A$.

\[\Rightarrow \dfrac{1-\dfrac{9}{16}}{1+\dfrac{9}{16}}=\dfrac{16}{25}-\dfrac{9}{25}\]

\[\Rightarrow \dfrac{7}{25}\]

\[\therefore \dfrac{1-{{\tan }^{2}}A}{1+{{\tan }^{2}}A}=\dfrac{7}{25}\]

R.H.S \[={{\cos }^{2}}A-{{\sin }^{2}}A\]

\[\Rightarrow {{\left( \dfrac{4}{5} \right)}^{2}}-{{\left( \dfrac{3}{5} \right)}^{2}}=\dfrac{16}{25}-\dfrac{9}{25}\]

$\therefore {{\cos }^{2}}A-{{\sin }^{2}}A=\dfrac{7}{25}$.

It shows that \[\text{L}\text{.H}\text{.S=R}\text{.H}\text{.S}\]

\[\therefore \dfrac{1-{{\tan }^{2}}A}{1+{{\tan }^{2}}A}={{\cos }^{2}}A-{{\sin }^{2}}A\]

4. In \[\Delta ABC\] Right Angles at \[B\], if \[A=\dfrac{1}{\sqrt{3}}\], then Find Value of the Following Equations:                                  

(i) \[\sin A\cos C+\cos A\sin C\]

Let \[BC=k\]and \[AB=\sqrt{3}k\]

Then, using Pythagoras theorem find $AC$

\[\Rightarrow \sqrt{{{(k)}^{2}}+{{(\sqrt{3}k)}^{2}}}\Rightarrow \sqrt{{{k}^{2}}+3{{k}^{2}}}\Rightarrow \sqrt{4{{k}^{2}}}\Rightarrow 2k\]

\[\therefore AC=2k\]

\[\sin A=\dfrac{BC}{AC}=\dfrac{k}{2k}=\dfrac{1}{2}\] and \[\cos A=\dfrac{AB}{AC}=\dfrac{\sqrt{3}k}{2k}=\dfrac{\sqrt{3}}{2}\]

For \[\angle C\], adjacent \[=BC\], opposite \[=AB\], and hypotenuse \[=AC\]

\[\sin C=\dfrac{AB}{AC}=\dfrac{\sqrt{3}k}{2k}=\dfrac{\sqrt{3}}{2}\] and \[\cos A=\dfrac{BC}{AC}=\dfrac{k}{2k}=\dfrac{1}{2}\]

Now find the values of the following equations,

\[\sin A\cos C+\cos A\sin C\]

\[\Rightarrow \dfrac{1}{2}\times \dfrac{1}{2}+\dfrac{\sqrt{3}}{2}\times \dfrac{\sqrt{3}}{2}\Rightarrow \dfrac{1}{4}+\dfrac{3}{4}\Rightarrow \dfrac{4}{4}\Rightarrow 1\]

\[\therefore \sin A\cos C+\cos A\sin C=1\]

(ii) \[\cos A\cos C-\sin A\sin C\]

\[\Rightarrow \dfrac{\sqrt{3}}{2}\times \dfrac{1}{2}-\dfrac{1}{2}\times \dfrac{\sqrt{3}}{2}\Rightarrow \dfrac{\sqrt{3}}{4}-\dfrac{\sqrt{3}}{4}\Rightarrow 0\]

\[\therefore \cos A\cos C-\sin A\sin C=0\]

5. In \[\Delta PQR\], right angled at \[Q\],\[PR+QR=25cm\] and \[PQ=5cm\]. Determine the Values of  \[\sin P,\cos P\]  and \[\tan P\].                                                                      

Given: In \[\Delta PQR\], right angled at \[Q\]

And \[PR+QR=25cm\], \[PQ=5cm\]

Let us take \[QR=xcm\] and \[PR=(25-x)cm\]

By using Pythagoras theorem, find the value of $x$.

\[R{{P}^{2}}=R{{Q}^{2}}+Q{{P}^{2}}\]

\[\Rightarrow {{(25-x)}^{2}}={{(x)}^{2}}+{{(5)}^{2}}\Rightarrow 625-50x+{{x}^{2}}={{x}^{2}}+25\]

\[\Rightarrow -50x=-600\Rightarrow x=12\]

Hence, \[RQ=12cm\]and \[RP=25-12=13cm\]

Now, find the values of \[\sin P,\cos P\]   and \[\tan P\].     

\[\therefore \sin P=\dfrac{RQ}{RP}=\dfrac{12}{13}\], \[\cos P=\dfrac{PQ}{RP}=\dfrac{5}{13}\] and \[\tan P=\dfrac{RQ}{PQ}=\dfrac{12}{5}\].

6. If \[\tan (A+B)=\sqrt{3}\] and \[\tan (A-B)=\dfrac{1}{\sqrt{3}}\]; \[{{0}^{\circ }}<A+B\le {{90}^{\circ }}\]; \[A>B\]. Find \[A\]and \[B\].  

Given : $\tan \left( A+B \right)=\sqrt{3}$ and $\tan \left( A-B \right)=\dfrac{1}{\sqrt{3}}$.

We know that,$\tan 60{}^\circ =\sqrt{3}$ and $\tan 30{}^\circ =\dfrac{1}{\sqrt{3}}$.

$\tan \left( A+B \right)=\tan 60{}^\circ $

$\Rightarrow A+B=60{}^\circ $          …… (1)

$\tan \left( A-B \right)=\tan 30{}^\circ $

$\Rightarrow A-B=30{}^\circ $          …… (2)

Adding equation (1) and (2). We get,

$\begin{align} & A+B+A-B=60{}^\circ +30{}^\circ \\ & \Rightarrow 2A=90{}^\circ \Rightarrow A=45{}^\circ \\ \end{align}$

$\therefore A=45{}^\circ $

Put $A=45{}^\circ $ in equation (1).

$A+B=60{}^\circ $

$\Rightarrow 45{}^\circ +B=60{}^\circ \Rightarrow B=60{}^\circ -45{}^\circ \Rightarrow B=15{}^\circ $

$\therefore B=15{}^\circ $

Hence, $A=45{}^\circ $ and $B=15{}^\circ $.

7. Choose the Correct Option. Justify Your Choice:   

(i) \[9{{\sec }^{2}}A-9{{\tan }^{2}}A\]\[=\]

\[1\] 

\[9\] 

\[8\] 

Ans: (B) $9$

\[9{{\sec }^{2}}A-9{{\tan }^{2}}A\]

\[\Rightarrow 9({{\sec }^{2}}A-{{\tan }^{2}}A)\Rightarrow 9\times 1\Rightarrow 9\]

(ii) \[(1+\tan \theta +\sec \theta )(1+\cot \theta -\cos ec\theta )\]\[=\]

\[0\] 

\[2\] 

none of these

Ans: (C) \[2\]

\[(1+\tan \theta +\sec \theta )(1+\cot \theta -\cos ec\theta )\]

\[\Rightarrow \left( 1+\dfrac{\sin \theta }{\cos \theta }+\dfrac{1}{\cos \theta } \right)\left( 1+\dfrac{\cos \theta }{\sin \theta }-\dfrac{1}{\sin \theta } \right)\Rightarrow \left( \dfrac{\cos \theta +\sin \theta +1}{\cos \theta } \right)\left( \dfrac{\sin \theta +\cos \theta -1}{\sin \theta } \right)\]

We know that, \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\]

\[\Rightarrow \dfrac{{{(\cos \theta +\sin \theta )}^{2}}-{{(1)}^{2}}}{\cos \theta .\sin \theta }\Rightarrow \dfrac{{{\cos }^{2}}\theta +{{\sin }^{2}}\theta +2\cos \theta \sin \theta -1}{\cos \theta .\sin \theta }\]        

\[\Rightarrow \dfrac{1+2\cos \theta \sin \theta -1}{\cos \theta .\sin \theta }\Rightarrow \dfrac{2\cos \theta \sin \theta }{\cos \theta .\sin \theta }\Rightarrow 2\]

(iii) \[(\sec A+\tan A)(1-\sin A)\]\[=\]

\[\sec A\] 

\[\sin A\] 

\[\cos ecA\] 

Ans : (D) \[\cos A\]

\[(\sec A+\tan A)(1-\sin A)\]

\[\Rightarrow \left( \dfrac{1}{\cos A}+\dfrac{\sin A}{\cos A} \right)(1-\sin A)\Rightarrow \left( \dfrac{1+\sin A}{\cos A} \right)(1-\sin A)\]

We know that, \[1-{{\sin }^{2}}A={{\cos }^{2}}A\]

\[\Rightarrow \dfrac{1-{{\sin }^{2}}A}{\cos A}\Rightarrow \dfrac{{{\cos }^{2}}A}{\cos A}\Rightarrow \cos A\]  

(iv) \[\dfrac{1+{{\tan }^{2}}A}{1+{{\cot }^{2}}A}=\]

\[{{\sec }^{2}}A\] 

\[-1\] 

\[{{\cot }^{2}}A\] 

Ans: (D) ${{\tan }^{2}}A$

\[\dfrac{1+{{\tan }^{2}}A}{1+{{\cot }^{2}}A}=\dfrac{{{\sec }^{2}}A-{{\tan }^{2}}A+{{\tan }^{2}}A}{\cos e{{c}^{2}}A-{{\cot }^{2}}A+{{\cot }^{2}}A}\]

\[\Rightarrow \dfrac{{{\sec }^{2}}A}{\cos e{{c}^{2}}A}\Rightarrow \dfrac{\dfac{rac{1}{{{\cos }^{2}}A}}{\dfrac{1}{{{\sin }^{2}}A}}\]

\[\Rightarrow \dfr{{\sin }^{2}}A}{{{\cos }^{2}}A}\Rightarrow {{\tan }^{2}}A\].

Long Answer Questions (4 Marks)

1. Express the Trigonometric Ratios \[\sin A,\sec A\] and \[\tan A\] in Terms of \[\cot A\].  

Find the value for \[\sin A\] in terms of \[\cot A\]

By using identity \[\cos e{{c}^{2}}A-{{\cot }^{2}}A=1\]

$\therefore \sin A=\dfrac{1}{\sqrt{1+{{\cot }^{2}}A}}$

Express the value of \[\sec A\] in terms of \[\cot A\]

By using identity \[{{\sec }^{2}}A-{{\tan }^{2}}A=1\]

\[\Rightarrow {{\sec }^{2}}A=\dfrac{1+{{\cot }^{2}}A}{{{\cot }^{2}}A}\]

\[\Rightarrow \sec A=\dfrac{\sqrt{1+{{\cot }^{2}}A}}{\cot A}\]

Express the value of \[\tan A\] in terms of \[\cot A\]

We know that, $\tan A=\dfrac{1}{\cot A}$

\[\therefore \tan A=\dfrac{1}{\cot A}\]

2. Write the Other Trigonometric Ratios of \[A\] in Terms of \[\sec A\] .                 

Express the value of \[\sin A\] in terms of \[\sec A\]

By using identity, \[{{\sin }^{2}}A+{{\cos }^{2}}A=1\]

\[\Rightarrow {{\sin }^{2}}A=1-{{\cos }^{2}}A\Rightarrow 1-\dfrac{1}{{{\sec }^{2}}A}\Rightarrow \dfrac{{{\sec }^{2}}A-1}{{{\sec }^{2}}A}\]

Express the value of \[\cos A\] in terms of \[\sec A\]

We know that, \[\cos A=\dfrac{1}{\sec A}\]

\[\therefore \cos A=\dfrac{1}{\sec A}\]

Express the value of \[\tan A\] in terms of \[\sec A\]

$\therefore \tan A=\sqrt{{{\sec }^{2}}A-1}$

Express the value of \[\cos ecA\] in terms of \[\sec A\]

We know that, $\operatorname{cosec}A=\dfrac{1}{\sin A}$

Substitute the value of \[\sin A\]

\[\Rightarrow \cos ecA=\dfrac{1}{\dfrac{\sqrt{{{\sec }^{2}}A-1}}{\sec A}}\Rightarrow \dfrac{\sec A}{\sqrt{{{\sec }^{2}}A-1}}\]

$\therefore \operatorname{cosec}A=\dfrac{\sec A}{\sqrt{{{\sec }^{2}}A-1}}$

Express the value of \[\cot A\] in terms of \[\sec A\]

We know that, \[\cot A=\dfrac{1}{\tan A}\]

Substitute the value of \[\tan A\]

$\therefore \cot A=\dfrac{1}{\sqrt{{{\sec }^{2}}A-1}}$

3. Evaluate the following equations:                                                                                                       

(i). \[\dfrac{{{\sin }^{2}}{{63}^{\circ }}+{{\sin }^{2}}{{27}^{\circ }}}{{{\cos }^{2}}{{17}^{\circ }}+{{\cos }^{2}}{{73}^{\circ }}}\]

We know that, \[\sin ({{90}^{\circ }}-\theta )=\cos \theta ,\,\cos ({{90}^{\circ }}-\theta )=\sin \theta \] and \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\]

\[\Rightarrow \dfrac{{{\sin }^{2}}{{63}^{\circ }}+{{\sin }^{2}}({{90}^{\circ }}-{{63}^{\circ }})}{{{\cos }^{2}}({{90}^{\circ }}-{{73}^{\circ }})+{{\cos }^{2}}{{73}^{\circ }}}\]         

\[\Rightarrow \dfrac{{{\sin }^{2}}{{63}^{\circ }}+{{\cos }^{2}}{{63}^{\circ }}}{{{\sin }^{2}}{{73}^{\circ }}+{{\cos }^{2}}{{73}^{\circ }}}\]                   

\[\Rightarrow \dfrac{1}{1}\Rightarrow 1\]

$\therefore \dfrac{{{\sin }^{2}}{{63}^{\circ }}+{{\sin }^{2}}{{27}^{\circ }}}{{{\cos }^{2}}{{17}^{\circ }}+{{\cos }^{2}}{{73}^{\circ }}}=1$

(ii). \[\sin {{25}^{\circ }}\cos {{65}^{\circ }}+\cos {{25}^{\circ }}\sin {{65}^{\circ }}\]

\[\Rightarrow \sin {{25}^{\circ }}\cos ({{90}^{\circ }}-{{25}^{\circ }})+\cos {{25}^{\circ }}\sin ({{90}^{\circ }}-{{25}^{\circ }})\]

\[\Rightarrow \sin {{25}^{\circ }}.\sin {{25}^{\circ }}+\cos {{25}^{\circ }}.\cos {{25}^{\circ }}\]                   

\[\Rightarrow {{\sin }^{2}}{{25}^{\circ }}+{{\cos }^{2}}{{25}^{\circ }}=1\]      

\[\therefore \sin {{25}^{\circ }}\cos {{65}^{\circ }}+\cos {{25}^{\circ }}\sin {{65}^{\circ }}=1\]

4. Choose the Correct Option. Justify Your Choice:  

\[\Rightarrow \dfrac{1-{{\sin }^{2}}A}{\cos A}\Rightarrow \dfrac{{{\cos }^{2}}A}{\cos A}\Rightarrow \cos A\] 

\[\Rightarrow \dfrac{{{\sec }^{2}}A}{\cos e{{c}^{2}}A}\Rightarrow \dfrac{\dfrac{1}{{{\cos }^{2}}A}}{\dfrac{1}{{{\sin }^{2}}A}}\]

\[\Rightarrow \dfrac{{{\sin }^{2}}A}{{{\cos }^{2}}A}\Rightarrow {{\tan }^{2}}A\].

5. Prove the Following Identities, Where the Angles Involved are Acute Angles for Which the Expressions are Defined:                      

(i). ${{(\operatorname{cosec}\theta -\cot \theta )}^{2}}=\dfrac{1-\cos \theta }{1+\cos \theta }$

Given: ${{(\operatorname{cosec}\theta -\cot \theta )}^{2}}=\dfrac{1-\cos \theta }{1+\cos \theta }$

We know that, \[{{(a-b)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab\], $\operatorname{cosec}\theta =\dfrac{1}{\sin \theta }$ and $\cot \theta =\dfrac{\cos \theta }{\sin \theta }$

Then, let us take left-hand side

\[\Rightarrow {{(\operatorname{cosec}\theta -\cot \theta )}^{2}}={{\operatorname{cosec}}^{2}}\theta +{{\cot }^{2}}\theta -2\operatorname{cosec}\theta \cot \theta \] 

\[\Rightarrow \dfrac{1}{{{\sin }^{2}}\theta }+\dfrac{{{\cos }^{2}}\theta }{{{\sin }^{2}}\theta }-2\times \dfrac{1}{\sin \theta }.\dfrac{\cos \theta }{\sin \theta }\Rightarrow \dfrac{1+{{\cos }^{2}}\theta }{{{\sin }^{2}}\theta }-\dfrac{2\cos \theta }{{{\sin }^{2}}\theta }\]

\[\Rightarrow \dfrac{1+{{\cos }^{2}}\theta -2\cos \theta }{{{\sin }^{2}}\theta }\Rightarrow \dfrac{{{(1-\cos \theta )}^{2}}}{{{\sin }^{2}}\theta }\]

\[\Rightarrow \dfrac{(1-\cos \theta )(1-\cos \theta )}{1-{{\cos }^{2}}\theta }\Rightarrow \dfrac{(1-\cos \theta )(1-\cos \theta )}{(1+\cos \theta )(1-\cos \theta )}\]

\[\Rightarrow \dfrac{1-\cos \theta }{1+\cos \theta }=\text{R}\text{.H}\text{.S}\]

$\therefore {{(\operatorname{cosec}\theta -\cot \theta )}^{2}}=\dfrac{1-\cos \theta }{1+\cos \theta }$

(ii) \[\dfrac{\cos A}{1+\sin A}+\dfrac{1+\sin A}{\cos A}=2\sec A\]

Given: \[\dfrac{\cos A}{1+\sin A}+\dfrac{1+\sin A}{\cos A}=2\sec A\]

\[\dfrac{\cos A}{1+\sin A}+\dfrac{1+\sin A}{\cos A}\Rightarrow \dfrac{{{\cos }^{2}}A+1+{{\sin }^{2}}A+2\sin A}{(1+\sin A)\cos A}\]

\[\Rightarrow \dfrac{{{\cos }^{2}}A+{{\sin }^{2}}A+1+2\sin A}{(1+\sin A)\cos A}\]      

\[\Rightarrow \dfrac{1+1+2\sin A}{(1+\sin A)\cos A}\]

\[\Rightarrow \dfrac{2+2\sin A}{(1+\sin A)\cos A}\]

\[\Rightarrow \dfrac{2(1+\sin A)}{(1+\sin A)\cos A}\]

\[\Rightarrow \dfrac{2}{\cos A}\Rightarrow 2\sec A=\text{R}\text{.H}\text{.S}\]

$\therefore \dfrac{\cos A}{1+\sin A}+\dfrac{1+\sin A}{\cos A}=2\sec A$

(iii). \[\dfrac{\tan \theta }{1-\cot \theta }+\dfrac{\cot \theta }{1-\tan \theta }=1+\sec \theta \operatorname{cosec}\theta \]

Given: \[\dfrac{\tan \theta }{1-\cot \theta }+\dfrac{\cot \theta }{1-\tan \theta }=1+\sec \theta \operatorname{cosec}\theta \]

We know that, \[{{a}^{3}}-{{b}^{3}}=(a-b)({{a}^{2}}+{{b}^{2}}+ab)\] and \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\]

Then, let us take L.H.S

\[\dfrac{\tan \theta }{1-\cot \theta }+\dfrac{\cot \theta }{1-\tan \theta }=\dfrac{\dfrac{\sin \theta }{\cos \theta }}{1-\dfrac{\cos \theta }{\sin \theta }}+\dfrac{\dfrac{\cos \theta }{\sin \theta }}{1-\dfrac{\sin \theta }{\cos \theta }}\]

\[\Rightarrow \dfrac{\sin \theta }{\cos \theta }\times \dfrac{\sin \theta }{\sin \theta -\cos \theta }+\dfrac{\cos \theta }{\sin \theta }\times \dfrac{\cos \theta }{\cos \theta -\sin \theta }\]

\[\Rightarrow \dfrac{{{\sin }^{2}}\theta }{\cos \theta (\sin \theta -\cos \theta )}-\dfrac{{{\cos }^{2}}\theta }{\sin \theta (\sin \theta -\cos \theta )}\]

\[\Rightarrow \dfrac{{{\sin }^{3}}\theta -{{\cos }^{3}}\theta }{\sin \theta \cos \theta (\sin \theta -\cos \theta )}\]                     

\[\Rightarrow \dfrac{1+\sin \theta \cos \theta }{\sin \theta \cos \theta }\]

\[\Rightarrow \dfrac{1}{\sin \theta \cos \theta }+1\]

\[\Rightarrow 1+\dfrac{1}{\sin \theta \cos \theta }\Rightarrow 1+\sec \theta \cos ec\theta \] = R.H.S

$\therefore \dfrac{\tan \theta }{1-\cot \theta }+\dfrac{\cot \theta }{1-\tan \theta }=1+\sec \theta \operatorname{cosec}\theta $

(iv) \[\dfrac{1+\sec A}{\sec A}=\dfrac{{{\sin }^{2}}A}{1-\cos A}\]

Given: \[\dfrac{1+\sec A}{\sec A}=\dfrac{{{\sin }^{2}}A}{1-\cos A}\]

Then, let us take L.H.S 

\[\dfrac{1+\sec A}{\sec A}=\dfrac{1+\dfrac{1}{\cos A}}{\dfrac{1}{\cos A}}\]

\[\Rightarrow \dfrac{\cos A+1}{\cos A}\times \dfrac{\cos A}{1}\Rightarrow 1+\cos A\]

\[\Rightarrow 1+\cos A\times \dfrac{1-\cos A}{1-\cos A}\Rightarrow \dfrac{1-{{\cos }^{2}}A}{1-\cos A}\]

\[\Rightarrow \dfrac{{{\sin }^{2}}A}{1-\cos A}=\text{R}\text{.H}\text{.S}\]

$\therefore \dfrac{1+\sec A}{\sec A}=\dfrac{{{\sin }^{2}}A}{1-\cos A}$

(v) \[\dfrac{\cos A-\sin A+1}{\cos A+\sin A-1}=\cos ecA+\cot A\], using the identity \[\cos e{{c}^{2}}A=1+{{\cot }^{2}}A\]

Given: \[\dfrac{\cos A-\sin A+1}{\cos A+\sin A-1}=\cos ecA+\cot A\]

We know that, \[\cos e{{c}^{2}}A=1+{{\cot }^{2}}A\]

\[\dfrac{\cos A-\sin A+1}{\cos A+\sin A-1}\]

Dividing all terms by \[\sin A\]

\[\Rightarrow \dfrac{\cot A-1+\cos ecA}{\cot A+1-\cos ecA}\]

\[\Rightarrow \dfrac{\cot A+\cos ecA-1}{\cot A-\cos ecA+1}\]

\[\Rightarrow \dfrac{(\cot A+\cos ecA)-(\cos e{{c}^{2}}A-{{\cot }^{2}}A)}{(1+\cot A-\cos ecA)}\]

\[\Rightarrow \dfrac{(\cot A+\cos ecA)+({{\cot }^{2}}A-\cos e{{c}^{2}}A)}{(1+\cot A-\cos ecA)}\]

\[\Rightarrow \dfrac{(\cot A+\cos ecA)(1+\cot A-\cos ecA)}{(1+\cot A-\cos ecA)}\]

\[\Rightarrow \cot A+\cos ecA=\text{R}\text{.H}\text{.S}\]

$\therefore \dfrac{\cos A-\sin A+1}{\cos A+\sin A-1}=\cos ecA+\cot A$

(vi) \[\sqrt{\dfrac{1+\sin A}{1-\sin A}}=\sec A+\tan A\]

Given: \[\sqrt{\dfrac{1+\sin A}{1-\sin A}}=\sec A+\tan A\]

We know that, \[1-{{\sin }^{2}}\theta ={{\cos }^{2}}\theta \] and \[(a+b)(a-b)={{a}^{2}}-{{b}^{2}}\]

\[\sqrt{\dfrac{1+\sin A}{1-\sin A}}\]

Let us take conjugate of the term. Then,

\[\Rightarrow \sqrt{\dfrac{1+\sin A}{1-\sin A}}\times \sqrt{\dfrac{1+\sin A}{1+\sin A}}\]

\[\Rightarrow \sqrt{\dfrac{{{(1+\sin A)}^{2}}}{1-{{\sin }^{2}}A}}\]

\[\Rightarrow \sqrt{\dfrac{{{\left( 1+\sin A \right)}^{2}}}{{{\operatorname{Cos}}^{2}}A}}\]

\[\Rightarrow \dfrac{1+\sin A}{\cos A}\Rightarrow \dfrac{1}{\cos A}+\dfrac{\sin A}{\cos A}\]                           

\[\Rightarrow \sec A+\tan A=\text{R}\text{.H}\text{.S}\]           

$\therefore \sqrt{\dfrac{1+\sin A}{1-\sin A}}=\sec A+\tan A$       

Hence proved.        

(vii) \[\dfrac{\sin \theta -2{{\sin }^{3}}\theta }{2{{\cos }^{3}}\theta -\cos \theta }=\tan \theta \]

Given: \[\dfrac{\sin \theta -2{{\sin }^{3}}\theta }{2{{\cos }^{3}}\theta -\cos \theta }=\tan \theta \]

We know that, \[1-{{\sin }^{2}}\theta ={{\cos }^{2}}\theta \]

\[\dfrac{\sin \theta -2{{\sin }^{3}}\theta }{2{{\cos }^{3}}\theta -\cos \theta }=\dfrac{\sin \theta (1-2{{\sin }^{2}}\theta )}{\cos \theta (2{{\cos }^{2}}\theta -1)}\]

\[\Rightarrow \dfrac{\sin \theta (1-2{{\sin }^{2}}\theta )}{\cos \theta \left[ 2(1-{{\sin }^{2}}\theta )-1 \right]}\]

\[\Rightarrow \dfrac{\sin \theta (1-2{{\sin }^{2}}\theta )}{\cos \theta (2-2{{\sin }^{2}}\theta -1)}\]

\[\Rightarrow \dfrac{\sin \theta (1-2{{\operatorname{Sin}}^{2}}\theta )}{\cos \theta (1-2{{\operatorname{Sin}}^{2}}\theta )}\]

\[\Rightarrow \dfrac{\sin \theta }{\cos \theta }=\tan \theta =\text{R}\text{.H}\text{.S}\]

$\therefore \dfrac{\sin \theta -2{{\sin }^{3}}\theta }{2{{\cos }^{3}}\theta -\cos \theta }=\tan \theta $

(viii) \[{{(\sin A+\cos ecA)}^{2}}+{{(\cos A+\sec A)}^{2}}=7+{{\tan }^{2}}A+{{\cot }^{2}}A\]

Given: \[{{(\sin A+\cos ecA)}^{2}}+{{(\cos A+\sec A)}^{2}}=7+{{\tan }^{2}}A+{{\cot }^{2}}A\]

We know that, \[\cos e{{c}^{2}}\theta =1+{{\cot }^{2}}\theta \] and \[{{\sec }^{2}}\theta =1+{{\tan }^{2}}\theta \]

\[{{(\sin A+\cos ecA)}^{2}}+{{(\cos A+\sec A)}^{2}}={{\left( \sin A+\dfrac{1}{\sin A} \right)}^{2}}+{{\left( \cos A+\dfrac{1}{\cos A} \right)}^{2}}\]

\[={{\sin }^{2}}A+\dfrac{1}{{{\sin }^{2}}A}+2\sin A.\dfrac{1}{\sin A}+{{\cos }^{2}}A+\dfrac{1}{{{\cos }^{2}}A}+2\cos A.\dfrac{1}{\cos A}\]

\[=2+2+{{\sin }^{2}}A+{{\cos }^{2}}A+\dfrac{1}{{{\sin }^{2}}A}+\dfrac{1}{{{\cos }^{2}}A}\]

\[=4+1+\dfrac{1}{{{\sin }^{2}}A}+\dfrac{1}{{{\cos }^{2}}A}\]

\[=5+\cos e{{c}^{2}}A+{{\sec }^{2}}A\]

\[=5+1+{{\cot }^{2}}A+1+{{\tan }^{2}}A\]

\[=7+{{\tan }^{2}}A+{{\cot }^{2}}A=\text{R}\text{.H}\text{.S}\]

$\therefore {{(\sin A+\cos ecA)}^{2}}+{{(\cos A+\sec A)}^{2}}=7+{{\tan }^{2}}A+{{\cot }^{2}}A$

(ix) \[(\cos ecA-\sin A)(\sec A-\cos A)=\dfrac{1}{\tan A+\cot A}\]

Given: \[(\cos ecA-\sin A)(\sec A-\cos A)=\dfrac{1}{\tan A+\cot A}\]

\[(\cos ecA-\sin A)(\sec A-\cos A)=\left( \dfrac{1}{\sin A}-\sin A \right)\left( \dfrac{1}{\cos A}-\cos A \right)\]

\[=\left( \dfrac{1-{{\sin }^{2}}A}{\sin A} \right)\left( \dfrac{1-{{\cos }^{2}}A}{\cos A} \right)\]

\[=\dfrac{{{\cos }^{2}}A}{\sin A}\times \dfrac{{{\sin }^{2}}A}{\cos A}\]

\[=\sin A.\cos A\]

\[=\dfrac{\sin A.\cos A}{{{\sin }^{2}}A+{{\cos }^{2}}A}\]

Dividing all the terms by \[\sin A.\cos A\]

\[=\dfrac{\dfrac{\sin A.\cos A}{\sin A.\cos A}}{\dfrac{{{\sin }^{2}}A}{\sin A.\cos A}+\dfrac{{{\cos }^{2}}A}{\sin A.\cos A}}\]

\[=\dfrac{1}{\dfrac{\sin A}{\cos A}+\dfrac{\cos A}{\sin A}}\]

\[=\dfrac{1}{\tan A+\cot A}=\text{R}\text{.H}\text{.S}\]

$\therefore (\cos ecA-\sin A)(\sec A-\cos A)=\dfrac{1}{\tan A+\cot A}$

(x) \[\left( \dfrac{1+{{\tan }^{2}}A}{1+{{\cot }^{2}}A} \right)={{\left( \dfrac{1-\tan A}{1-\cot A} \right)}^{2}}={{\tan }^{2}}A\]

Given:\[\left( \dfrac{1+{{\tan }^{2}}A}{1+{{\cot }^{2}}A} \right)={{\left( \dfrac{1-\tan A}{1-\cot A} \right)}^{2}}={{\tan }^{2}}A\]

We know that, \[1+{{\tan }^{2}}\theta ={{\sec }^{2}}\theta \] and \[1+{{\cot }^{2}}\theta =\cos e{{c}^{2}}A\]

\[\left( \dfrac{1+{{\tan }^{2}}A}{1+{{\cot }^{2}}A} \right)=\dfrac{{{\sec }^{2}}A}{\cos e{{c}^{2}}A}\]

\[=\dfrac{1}{{{\cos }^{2}}A}\times \dfrac{{{\sin }^{2}}A}{1}\]

\[={{\tan }^{2}}A=\text{R}\text{.H}\text{.S}\]

Now, prove the Middle side 

\[{{\left( \dfrac{1-\tan A}{1-\cot A} \right)}^{2}}={{\left( \dfrac{1-\tan A}{1-\dfrac{1}{\tan A}} \right)}^{2}}\]

\[={{\left( \dfrac{1-\tan A}{\dfrac{\tan A-1}{\tan A}} \right)}^{2}}\]

\[=\left( \dfrac{1-\tan A}{\dfrac{-(1-\tan A)}{\tan A}} \right)\]

\[={{(-\tan A)}^{2}}\]

$\therefore \left( \dfrac{1+{{\tan }^{2}}A}{1+{{\cot }^{2}}A} \right)={{\left( \dfrac{1-\tan A}{1-\cot A} \right)}^{2}}={{\tan }^{2}}A$

6. Use Euclid’s Division Algorithm to Find the \[\text{HCF}\] of:                                     

(i) \[\mathbf{135}\] and \[\mathbf{225}\]

Given: \[135\] and \[225\]

We have \[225>135\],

So, we have to apply the division lemma to \[225\] and \[135\] to obtain

\[225=135\times 1+90\]

Here remainder\[90\ne 0\], again we are applying the division lemma to \[135\] and \[90\] to obtain 

\[135=90\times 1+45\]

Again, the remainder\[45\ne 0\], then apply the division lemma to obtain 

\[90=2\times 45+0\]

Since now we got remainder as zero. Here, the process get stops. 

The divisor at this stage is \[45\]

Therefore, the HCF of \[135\] and \[225\] is \[45\].

(ii) \[\mathbf{196}\] and\[~\mathbf{38220}\] 

Given: \[196\] and \[38220\]

We have \[38220>196\], 

So, we have to apply the division lemma to \[38220\] and \[196\] to obtain 

\[38220=196\times 195+0\]

Since we get the remainder as zero, the process stops here. 

The divisor at this stage is \[196\], 

Therefore, HCF of \[196\] and \[38220\] is \[196\].  

(iii) \[~\mathbf{867}\] and \[\mathbf{255}\]

Given: \[867\] and \[255\]

We have 867 > 255, 

So, we have to apply the division lemma to \[867\]and \[255\] to obtain 

\[867=255\times 3+102\]

Here remainder \[102\ne 0\], again apply the division lemma to\[255\] and \[102\] to obtain 

\[255=102\times 2\text{ }51\]

Again, remainder \[51\ne 0\], again apply the division lemma to \[102\] and \[51\] to obtain 

\[102=51\times 2+0\]

The divisor at this stage is \[51\], 

Therefore, HCF of \[867\] and \[255\] is \[51\].

7. Evaluate the Following Equations:                                                                                                              

i. \[\sin {{60}^{\circ }}\cos {{30}^{\circ }}+\sin {{30}^{\circ }}\cos {{60}^{\circ }}\]

Given: \[\sin {{60}^{\circ }}\cos {{30}^{\circ }}+\sin {{30}^{\circ }}\cos {{60}^{\circ }}\]

We know that, $\sin 60{}^\circ =\dfrac{\sqrt{3}}{2},\,\sin 30{}^\circ =\dfrac{1}{2}$ and $\cos 60{}^\circ =\dfrac{1}{2},\,\cos 30{}^\circ =\dfrac{\sqrt{3}}{2}$

\[\sin {{60}^{\circ }}\cos {{30}^{\circ }}+\sin {{30}^{\circ }}\cos {{60}^{\circ }}\]

\[=\dfrac{\sqrt{3}}{2}\times \dfrac{\sqrt{3}}{2}+\dfrac{1}{2}\times \dfrac{1}{2}\]

\[=\dfrac{3}{4}+\dfrac{1}{4}\]

\[=\dfrac{4}{4}=1\]

\[\therefore \sin {{60}^{\circ }}\cos {{30}^{\circ }}+\sin {{30}^{\circ }}\cos {{60}^{\circ }}=1\]

ii. \[2{{\tan }^{2}}{{45}^{\circ }}+{{\cos }^{2}}{{30}^{\circ }}-{{\sin }^{2}}{{60}^{\circ }}\]

Given: \[2{{\tan }^{2}}{{45}^{\circ }}+{{\cos }^{2}}{{30}^{\circ }}-{{\sin }^{2}}{{60}^{\circ }}\]

We know that, $\tan 45{}^\circ =1$, $\sin 60{}^\circ =\dfrac{\sqrt{3}}{2}$ and $\cos 30{}^\circ =\dfrac{\sqrt{3}}{2}$

Then, \[2{{\tan }^{2}}{{45}^{\circ }}+{{\cos }^{2}}{{30}^{\circ }}-{{\sin }^{2}}{{60}^{\circ }}\]

\[=2{{(1)}^{2}}+{{\left( \dfrac{\sqrt{3}}{2} \right)}^{2}}-{{\left( \dfrac{\sqrt{3}}{2} \right)}^{2}}\]

\[=2+\dfrac{3}{4}-\dfrac{3}{4}=2\]

$\therefore 2{{\tan }^{2}}{{45}^{\circ }}+{{\cos }^{2}}{{30}^{\circ }}-{{\sin }^{2}}{{60}^{\circ }}=2$

iii. \[\dfrac{\cos {{45}^{\circ }}}{\sec {{30}^{\circ }}+\cos ec{{30}^{\circ }}}\]

Given: \[\dfrac{\cos {{45}^{\circ }}}{\sec {{30}^{\circ }}+\cos ec{{30}^{\circ }}}\]

We know that, $\cos 45{}^\circ =\dfrac{1}{\sqrt{2}},\,\sec 30{}^\circ =\dfrac{2}{\sqrt{3}}$ and $\operatorname{cosec}30{}^\circ =2$

\[=\dfrac{\dfrac{1}{\sqrt{2}}}{\dfrac{2}{\sqrt{3}}+2}\]

\[=\dfrac{\dfrac{1}{\sqrt{2}}}{\dfrac{2+2\sqrt{3}}{\sqrt{3}}}\]

\[=\dfrac{1}{\sqrt{2}}\times \dfrac{\sqrt{3}}{2+2\sqrt{3}}\]

\[=\dfrac{\sqrt{3}}{\sqrt{2}\times 2(\sqrt{3}+1)}\]

\[=\dfrac{\sqrt{3}}{\sqrt{2}\times 2(\sqrt{3}+1)}\times \dfrac{\sqrt{3}-1}{\sqrt{3}-1}\]

\[=\dfrac{\sqrt{3}\left( \sqrt{3}-1 \right)}{\sqrt{2}\times 2(3-1)}\]

\[=\dfrac{\sqrt{3}(\sqrt{3}-1)}{4\sqrt{2}}\times \dfrac{\sqrt{2}}{\sqrt{2}}\]\[\]

\[=\dfrac{3\sqrt{2}-\sqrt{6}}{8}\]

$\therefore \dfrac{\cos {{45}^{\circ }}}{\sec {{30}^{\circ }}+\cos ec{{30}^{\circ }}}=\dfrac{3\sqrt{2}-\sqrt{6}}{8}$

(iv) \[\dfrac{\sin {{30}^{\circ }}+\tan {{45}^{\circ }}-\cos ec{{60}^{\circ }}}{\sec {{30}^{\circ }}+\cos {{60}^{\circ }}+\cot {{45}^{\circ }}}\]

Given: \[\dfrac{\sin {{30}^{\circ }}+\tan {{45}^{\circ }}-\cos ec{{60}^{\circ }}}{\sec {{30}^{\circ }}+\cos {{60}^{\circ }}+\cot {{45}^{\circ }}}\]

We know that,  $\sin 30{}^\circ =\dfrac{1}{2},\,\tan 45{}^\circ =1,\,\operatorname{cosec}60{}^\circ =\dfrac{2}{\sqrt{3}},\,\sec 30{}^\circ =\dfrac{2}{\sqrt{3}},\,\cos 60{}^\circ =\dfrac{1}{2}$ and $\cot 45{}^\circ =1$.

Then, \[\dfrac{\sin {{30}^{\circ }}+\tan {{45}^{\circ }}-\cos ec{{60}^{\circ }}}{\sec {{30}^{\circ }}+\cos {{60}^{\circ }}+\cot {{45}^{\circ }}}\]

$=\dfrac{{1}/{2}\;+1-{2}/{\sqrt{3}}\;}{{2}/{\sqrt{3}}\;+{1}/{2}\;+1}$

\[=\dfrac{\dfrac{\sqrt{3}+2\sqrt{3}-4}{2\sqrt{3}}}{\dfrac{4+\sqrt{3}+2\sqrt{3}}{2\sqrt{3}}}\]

\[=\dfrac{3\sqrt{3}-4}{3\sqrt{3}+4}\]

\[=\dfrac{3\sqrt{3}-4}{3\sqrt{3}+4}\times \dfrac{3\sqrt{3}-4}{3\sqrt{3}-4}\]

\[=\dfrac{27+16-24\sqrt{3}}{27-16}\]

\[=\dfrac{43-24\sqrt{3}}{11}\]

$\therefore \dfrac{\sin {{30}^{\circ }}+\tan {{45}^{\circ }}-\cos ec{{60}^{\circ }}}{\sec {{30}^{\circ }}+\cos {{60}^{\circ }}+\cot {{45}^{\circ }}}=\dfrac{43-24\sqrt{3}}{11}$

(v) \[\dfrac{5{{\cos }^{2}}{{60}^{\circ }}+4{{\sin }^{2}}{{30}^{\circ }}-{{\tan }^{2}}{{45}^{\circ }}}{{{\sin }^{2}}{{30}^{\circ }}+{{\cos }^{2}}{{30}^{\circ }}}\]

Given: \[\dfrac{5{{\cos }^{2}}{{60}^{\circ }}+4{{\sin }^{2}}{{30}^{\circ }}-{{\tan }^{2}}{{45}^{\circ }}}{{{\sin }^{2}}{{30}^{\circ }}+{{\cos }^{2}}{{30}^{\circ }}}\]. Then,

\[=\dfrac{5{{\left( \dfrac{1}{2} \right)}^{2}}+{{\left( \dfrac{3}{\sqrt{3}} \right)}^{2}}-{{(1)}^{2}}}{{{\left( \dfrac{1}{2} \right)}^{2}}+{{\left( \dfrac{\sqrt{3}}{2} \right)}^{2}}}\]

\[=\dfrac{5\times \dfrac{1}{4}+4\times \dfrac{4}{3}-1}{\dfrac{1}{4}+\dfrac{3}{4}}\]

\[=\dfrac{\dfrac{1}{12}\times 67}{\dfrac{4}{4}}\]

\[=\dfrac{67}{12}\]

$\therefore \dfrac{5{{\cos }^{2}}{{60}^{\circ }}+4{{\sin }^{2}}{{30}^{\circ }}-{{\tan }^{2}}{{45}^{\circ }}}{{{\sin }^{2}}{{30}^{\circ }}+{{\cos }^{2}}{{30}^{\circ }}}=\dfrac{67}{12}$

8. Prove the Following Identities, Where the Angles Involved are Acute Angles for Which the Expressions are Defined:                                      

(i) ${{(\operatorname{cosec}\theta -\cot \theta )}^{2}}=\dfrac{1-\cos \theta  }{1+\cos \theta }$

(iii) \[\dfrac{\tan \theta }{1-\cot \theta }+\dfrac{\cot \theta }{1-\tan \theta }=1+\sec \theta \operatorname{cosec}\theta \]

Given: \[\left( \dfrac{1+{{\tan }^{2}}A}{1+{{\cot }^{2}}A} \right)={{\left( \dfrac{1-\tan A}{1-\cot A} \right)}^{2}}={{\tan }^{2}}A\]

Key Topics Covered in Chapter 8

Chapter 8 of Class 10 Mathematics consists of the discussion of the following key concepts.

Basic Trigonometry

Opposite & Adjacent Sides In A Right-Angled Triangle

Basic Trigonometric Ratios

Standard Values Of Trigonometric Ratios 

Complementary Trigonometric Ratios 

Significance of Important Questions of Trigonometry Class 10

Trigonometry is one of the most significant parts of Important Questions for Class 10 Maths Chapter 8. This chapter revolves around the memorization and conceptual understanding of the user’s problem-solving ability of a given space, preferably a triangle. These Class 10 Maths Trigonometry Important Questions help students to have a better understanding and application of trigonometry in the real world and clear the base around its introductory stage, to help students retain its benefits in the long run.

Did You Know?

Trigonometry has different applications and a few of them are mentioned below:

It is used in cartography for the creation of maps.

It has applications in the aviation industry and satellite systems.

It is also used to describe light and sound waves. 

This was the complete discussion on CBSE Class 10 Maths Chapter 8 Introduction to Trigonometry important questions. If you are a Class 10 student we highly recommend downloading and practising the important questions on trigonometry. Practising these important questions will not only help you understand the concepts but will also help you in analysing the important topics and exam patterns. 

We wish you all the very best! Be exam ready with Vedantu!

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FAQs on Important Questions for CBSE Class 10 Maths Chapter 8 Introduction to Trigonometry

1. Is Studying Chapter 8 Maths Class 10 Important Questions Helpful in Preparing for the Exams?

Studying Trigonometry Important Questions Class 10 helps you uncover the different aspects of trigonometry and teach the subject at its core. The pool of experienced teachers who have a complete idea of the marking schematics and all the necessary questions that can come in the exams are focused on providing the students with a complete kit that helps them secure maximum marks in their upcoming exams. Further, the step by step explanation of practical problems is what set our solutions apart, making it completely reliable a solution to start preparing for your exams.

2. What are the important questions in Chapter 8 Trigonometry of Class 10 Maths?

Class 10 Maths Chapter 8 is related to trigonometry and its functions. Important questions given on Vedantu for trigonometry Class 10 Chapter 8 can help students to prepare for their exams. Important questions are given according to the latest syllabus and pattern of CBSE. All questions are prepared using sample papers, previous year papers, and the latest trends in the CBSE board exams. Students can use important questions in trigonometry Class 10 to prepare for their final board exams. These solutions are available on Vedantu's official website(vedantu.com) and mobile app free of cost.

3. What is the easiest way to solve Chapter 8 Trigonometry of Class 10 Maths questions?

Students have to understand the basic concepts of Trigonometry for solving Class 10 Maths Chapter 8 questions. They have to memorize the different identities properly for solving the NCERT questions quickly. Students can also refer to the important questions in Class 10 Maths Chapter 8 available for free on Vedantu to understand the basic concepts and formulas used for solving trigonometry easily.  They can visit Vedantu to download important questions for Class 10 Maths Chapter 8 for regular practice.

4. How many exercises are there in Chapter 8 Trigonometry of Class 10 Maths?

In Trigonometry Class 10 Maths Chapter 8, there are four exercises. All exercises are based on different concepts related to trigonometry. Students need to solve all questions given in four exercises in the NCERT book to understand the concepts of trigonometry and for scoring high marks in Class 10 maths board exams. Students can practice solving trigonometry from the NCERT Solutions available on Vedantu in the offline and online mode. They can download the NCERT solutions for Class 10 Maths Chapter 8 on computers to work offline.

5. What are the different topics studied in Chapter 8 Trigonometry of Class 10 Maths?

The different topics studied in Class 10 Maths Chapter 10 include an introduction to trigonometry,  information about trigonometric ratios, trigonometric ratios of specific angles, trigonometric ratios of complementary angles, and trigonometric identities. Students will learn all topics step by step in each exercise of the chapter. They can practice NCERT Solutions given on different topics of trigonometry on Vedantu. NCERT Solutions for trigonometry Class 10 are prepared by experts to help students understand the concepts and score high marks. 

CBSE Class 10 Maths Important Questions

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