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Class 11 Mathematics Case Study Questions

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The rationale behind teaching Mathematics

The general rationale to teach Mathematics at the senior secondary level is to assist students:

• In knowledge acquisition and cognitive understanding of basic ideas, words, principles, symbols, and mastery of underlying processes and abilities, notably through motivation and visualization.
• To experience the flow of arguments while demonstrating a point or addressing an issue.
• To use the information and skills gained to address issues using several methods wherever possible.
• To cultivate a good mentality in order to think, evaluate, and explain coherently.
• To spark interest in the subject by taking part in relevant tournaments.
• To familiarise pupils with many areas of mathematics utilized in daily life.
• To pique students’ interest in studying mathematics as a discipline.

Case studies in Class 11 Mathematics

A case study in mathematics is a comprehensive examination of a specific mathematical topic or scenario. Case studies are frequently used to investigate the link between theory and practise, as well as the connections between different fields of mathematics. A case study will frequently focus on a specific topic or circumstance and will investigate it using a range of methodologies. These approaches may incorporate algebraic, geometric, and/or statistical analysis.

Sample Class 11 Mathematics case study questions

When it comes to preparing for Class 11 Mathematics, one of the best things Class 11 Mathematics students can do is to look at some Class 11 Mathematics sample case study questions. Class 11 Mathematics sample case study questions will give you a good idea of the types of Class 11 Mathematics sample case study questions that will be asked in the exam and help you to prepare more effectively.

Looking at sample questions is also a good way to identify any areas of weakness in your knowledge. If you find that you struggle with a particular topic, you can then focus your revision on that area.

myCBSEguide offers ample Class 11 Mathematics case study questions, so there is no excuse. With a little bit of preparation, Class 11 Mathematics students can boost their chances of getting the grade they deserve.

Some samples of Class 11 Mathematics case study questions are as follows:

Class 11 Mathematics case study question 1

• 9 km and 13 km
• 9.8 km and 13.8 km
• 9.5 km and 13.5 km
• 10 km and 14 km
• x  ≤   −1913
• x <  −1613
• −1613  < x <  −1913
• There are no solution.
• y  ≤   12 x+2
• y >  12 x+2
• y  ≥   12 x+2
• y <  12 x+2

Answer Key:

• (b) 9.8 km and 13.8 km
• (a) −1913   ≤  x 
• (b)  y >  12 x+2
• (d) (-5, 5)

Class 11 Mathematics case study question 2

• 2 C 1 × 13 C 10
• 2 C 1 × 10 C 13
• 1 C 2 × 13 C 10
• 2 C 10 × 13 C 10
• 6 C 2​ × 3 C 4   × 11 C 5 ​
• 6 C 2​ × 3 C 4   × 11 C 5
• 6 C 2​ × 3 C 5 × 11 C 4 ​
• 6 C 2 ​  ×   3 C 1 ​  × 11 C 5 ​
• (b) (13) 4  ways
• (c) 2860 ways.

Class 11 Mathematics case study question 3

Read the Case study given below and attempt any 4 sub parts: Father of Ashok is a builder, He planned a 12 story building in Gurgaon sector 5. For this, he bought a plot of 500 square yards at the rate of Rs 1000 /yard². The builder planned ground floor of 5 m height, first floor of 4.75 m and so on each floor is 0.25 m less than its previous floor.

Class 11 Mathematics case study question 4

Read the Case study given below and attempt any 4 sub parts: villages of Shanu and Arun’s are 50km apart and are situated on Delhi Agra highway as shown in the following picture. Another highway YY’ crosses Agra Delhi highway at O(0,0). A small local road PQ crosses both the highways at pints A and B such that OA=10 km and OB =12 km. Also, the villages of Barun and Jeetu are on the smaller high way YY’. Barun’s village B is 12km from O and that of Jeetu is 15 km from O.

Now answer the following questions:

• 5x + 6y = 60
• 6x + 5y = 60
• (a) (10, 0)
• (b) 6x + 5y = 60
• (b) 60/√ 61 km
• (d) 2√61 km

A peek at the Class 11 Mathematics curriculum

The Mathematics Syllabus has evolved over time in response to the subject’s expansion and developing societal requirements. The Senior Secondary stage serves as a springboard for students to pursue higher academic education in Mathematics or professional subjects such as Engineering, Physical and Biological Science, Commerce, or Computer Applications. The current updated curriculum has been prepared in compliance with the National Curriculum Framework 2005 and the instructions provided by the Focus Group on Teaching Mathematics 2005 in order to satisfy the rising demands of all student groups. Greater focus has been placed on the application of various principles by motivating the themes from real-life events and other subject areas.

Class 11 Mathematics (Code No. 041)

Design of Class 11 Mathematics exam paper

CBSE Class 11 mathematics question paper is designed to assess students’ understanding of the subject’s essential concepts. Class 11 mathematics question paper will assess their problem-solving and analytical abilities. Before beginning their test preparations, students in Class 11 maths should properly review the question paper format. This will assist Class 11 mathematics students in better understanding the paper and achieving optimum scores. Refer to the Class 11 Mathematics question paper design provided.

 Class 11 Mathematics Question Paper Design

• No chapter-wise weightage. Care to be taken to cover all the chapters.
• Suitable internal variations may be made for generating various templates keeping the overall weightage to different forms of questions and typology of questions the same.  

Choice(s): There will be no overall choice in the question paper. However, 33% of internal choices will be given in all the sections.

  Prescribed Books:

• Mathematics Textbook for Class XI, NCERT Publications
• Mathematics Exemplar Problem for Class XI, Published by NCERT
• Mathematics Lab Manual class XI, published by NCERT

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CBSE Case Study Questions for Class 11 Maths Sets Free PDF

Mere Bacchon, you must practice the CBSE Case Study Questions Class 11 Maths Sets 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 11 Maths Sets PDF

Checkout our case study questions for other chapters.

• Chapter 2 Relations and Functions Case Study Questions
• Chapter 3 Trigonometric Functions Case Study Questions
• Chapter 4 Principle of Mathematical Induction Case Study Questions
• Chapter 5 Complex Numbers and Quadratic Equations Case Study Questions

How should I study for my upcoming exams?

First, learn to sit for at least 2 hours at a stretch

Solve every question of NCERT by hand, without looking at the solution.

Solve NCERT Exemplar (if available)

Sit through chapter wise FULLY INVIGILATED TESTS

Practice MCQ Questions (Very Important)

Practice Assertion Reason & Case Study Based Questions

Sit through FULLY INVIGILATED TESTS involving MCQs. Assertion reason & Case Study Based Questions

After Completing everything mentioned above, Sit for atleast 6 full syllabus TESTS.

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CBSE Class 11 Maths – Chapter 3 Trigonometric Functions- Study Materials

NCERT Solutions Class 11 All Subjects Sample Papers Past Years Papers

Sets : Notes and Study Materials -pdf

• Concepts of  Trigonometric Functions
• Trigonometric Functions Master File
• Trigonometric Functions Revision Notes
• R D Sharma Solution of Trigonometric Functions
• NCERT Solution  Trigonometric Functions
• NCERT  Exemplar Solution Trigonometric Functions
• Trigonometric Functions : Solved Example 1

CBSE Class 11 Maths Notes Chapter 3 Trigonometric Functions

Angle Angle is a measure of rotation of a given ray about its initial point. The original ray is called the initial side and the final position of ray after rotation is called terminal side of the angle. The point of rotation is called vertex. If the direction of rotation is anti-clockwise, the angle is said to be positive and if the direction of rotation is clockwise, then the angle is negative.

Measuring Angles There are two systems of measuring angles Sexagesimal system (degree measure): If a rotation from the initial side to terminal side is  ( 1 360 ) t h  of a revolution, the angle is said to have a measure of one degree, written as 1°. One sixtieth of a degree is called a minute, written as 1′ and one-sixtieth of a minute is called a second, written as 1″ Thus, 1° = 60′ and 1′ = 60″

Circular system (radian measure):  A radian is an angle subtended at the centre of a circle by an arc, whose length is equal to the radius of the circle. We denote 1 radian by 1°.

Relation Between Radian and Degree We know that a complete circle subtends at its centre an angle whose measure is 2π radians as well as 360°. 2π radian = 360°. Hence, π radian = 180° or 1 radian = 57° 16′ 21″ (approx) 1 degree = 0.01746 radian

Six Fundamental Trigonometric Identities

• sinx =  1 c o s e c x
• cos x =  1 s e c x
• tan x =  1 c o t x
• sin 2  x + cos 2  x = 1
• 1 + tan 2 x = sec 2  x
• 1 + cot 2  x = cosec 2  x

Trigonometric Functions – Class 11 Maths Notes

Trigonometric ratios are defined for acute angles as the ratio of the sides of a right angled triangle. The extension of trigonometric ratios to any angle in terms of radian measure (real number) are called trigonometric function. The signs of trigonometric function in different quadrants have been given in following table.

Domain and Range of Trigonometric Functions

Sine, Cosine, and Tangent of Some Angles Less Than 90°

Allied or Related Angles The angles  n π 2 ± θ  are called allied or related angle and θ ± n × (2π) are called coterminal angles. For general reduction, we have following rules, the value of trigonometric function for ( n π 2 ± θ ) is numerically equal to

• the value of the same function, if n is an even integer with the algebraic sign of the function as per the quadrant in which angle lies.
• the corresponding co-function of θ, if n is an odd integer with the algebraic sign of the function for the quadrant in which it lies, here sine and cosine, tan and cot, sec and cosec are cofunctions of each other.

Functions of Negative Angles

For any acute angle of θ. We have,

• sin(-θ) = – sinθ
• cos (-θ) = cosθ
• tan (-θ) = – tanθ
• cot (-θ) = – cotθ
• sec (-θ) = secθ
• cosec (-θ) = – cosecθ

Some Formulae Regarding Compound Angles

An angle made up of the sum or difference of two or more angles is called compound angles. The basic results in direction are called trigonometric identities as given below: (i) sin (x + y) = sin x cos y + cos x sin y (ii) sin (x – y) = sin x cos y – cos x sin y (iii) cos (x + y) = cos x cos y – sin x sin y (iv) cos (x – y) = cos x cos y + sin x sin y

(ix) sin(x + y) sin (x – y) = sin 2  x – sin 2  y = cos 2  y – cos 2  x (x) cos (x + y) cos (x – y) = cos 2  x – sin 2  y = cos 2  y – sin 2  x

Transformation Formulae

• 2 sin x cos y = sin (x + y) + sin (x – y)
• 2 cos x sin y = sin (x + y) – sin (x – y)
• 2 cos x cos y = cos (x + y) + cos (x – y)
• 2 sin x sin y = cos (x – y) – cos (x + y)
• sin x + sin y = 2 sin( x + y 2 ) cos( x − y 2 )
• sin x – sin y = 2 cos( x + y 2 ) sin( x − y 2 )
• cos x + cos y = 2 cos( x + y 2 ) cos( x − y 2 )
• cos x – cos y = -2 sin( x + y 2 ) sin( x − y 2 )

Trigonometric Ratios of Multiple Angles

Product of Trigonometric Ratios

• sin x sin (60° – x) sin (60° + x) =  1 4  sin 3x
• cos x cos (60° – x) cos (60° + x) =  1 4  cos 3x
• tan x tan (60° – x) tan (60° + x) = tan 3x
• cos 36° cos 72° =  1 4
• cos x . cos 2x . cos 2 2 x . cos 2 3 x … cos 2 n-1  =  s i n 2 n x 2 n s i n x

Sum of Trigonometric Ratio, if Angles are in A.P.

Trigonometric Equations Equation which involves trigonometric functions of unknown angles is known as the trigonometric equation.

Solution of a Trigonometric Equation A solution of a trigonometric equation is the value of the unknown angle that satisfies the equation. A trigonometric equation may have an infinite number of solutions.

Principal Solution The solutions of a trigonometric equation for which 0 ≤ x ≤ 2π are called principal solutions.

General Solutions A solution of a trigonometric equation, involving ‘n’ which gives all solution of a trigonometric equation is called the general solutions.

General Solutions of Trigonometric Equation

• sin x = 0 ⇔ x = nπ, n ∈ Z
• cos x = 0 ⇔ x = (2n + 1)  π 2  , n ∈ Z
• tan x = 0 ⇔ x = nπ, n ∈ Z
• sin x = sin y ⇔ x = nπ + (-1) n  y, n ∈ Z
• cos x = cos y ⇔ x = 2nπ ± y, n ∈ Z
• tan x = tan y ⇔ x = nπ ± y, n ∈ Z
• sin 2  x = sin 2  y ⇔ x = nπ ± y, n ∈ Z
• cos 2  x = cos 2  y ⇔ x = nπ ± y, n ∈ Z
• tan 2  x = tan 2  y ⇔ x = nπ ± y, n ∈ Z

Basic Rules of Triangle

In a triangle ABC, the angles are denoted by capital letters A, B and C and the lengths of sides of opposite to these angles are denoted by small letters a, b and c, respectively. Sine Rule s i n A a = s i n B b = s i n C c

Cosine Rule a 2  = b 2  + c 2  – 2bc cos A b 2  = c 2  + a 2  – 2ac cos B c 2  = a 2  + b 2  – 2ab cos C

Projection Rule a = b cos C + c cos B b = c cos A + a cos C c = a cos B + b cos A

Trigonometric Functions Class 11 MCQs Questions with Answers

Question 1. The value of cos² x + cos² y – 2cos x × cos y × cos (x + y) is (a) sin (x + y) (b) sin² (x + y) (c) sin³ (x + y) (d) sin 4 (x + y)

Answer: (b) sin² (x + y) Hint: cos² x + cos² y – 2cos x × cos y × cos(x + y) {since cos(x + y) = cos x × cos y – sin x × sin y } = cos² x + cos² y – 2cos x × cos y × (cos x × cos y – sin x × sin y) = cos² x + cos² y – 2cos² x × cos² y + 2cos x × cos y × sin x × sin y = cos² x + cos² y – cos² x × cos² y – cos² x × cos² y + 2cos x × cos y × sin x × sin y = (cos² x – cos² x × cos² y) + (cos² y – cos² x × cos² y) + 2cos x × cos y × sin x × sin y = cos² x(1- cos² y) + cos² y(1 – cos² x) + 2cos x × cos y × sin x × sin y = sin² y × cos² x + sin² x × cos² y + 2cos x × cos y × sin x × sin y (since sin² x + cos² x = 1 ) = sin² x × cos² y + sin² y × cos² x + 2cos x × cos y × sin x × sin y = (sin x × cos y)² + (sin y × cos x)² + 2cos x × cos y × sin x × sin y = (sin x × cos y + sin y × cos x)² = {sin (x + y)}² = sin² (x + y)

Question 2. If a×cos x + b × cos x = c, then the value of (a × sin x – b²cos x)² is (a) a² + b² + c² (b) a² – b² – c² (c) a² – b² + c² (d) a² + b² – c²

Answer: (d) a² + b² – c² Hint: We have (a×cos x + b × sin x)² + (a × sin x – b × cos x)² = a² + b² ⇒ c² + (a × sin x – b × cos x)² = a² + b² ⇒ (a × sin x – b × cos x)² = a² + b² – c²

Question 3. If cos a + 2cos b + cos c = 2 then a, b, c are in (a) 2b = a + c (b) b² = a × c (c) a = b = c (d) None of these

Answer: (a) 2b = a + c Hint: Given, cos A + 2 cos B + cos C = 2 ⇒ cos A + cos C = 2(1 – cos B) ⇒ 2 cos((A + C)/2) × cos((A-C)/2 = 4 sin²(B/2) ⇒ 2 sin(B/2)cos((A-C)/2) = 4sin² (B/2) ⇒ cos((A-C)/2) = 2sin (B/2) ⇒ cos((A-C)/2) = 2cos((A+C)/2) ⇒ cos((A-C)/2) – cos((A+C)/2) = cos((A+C)/2) ⇒ 2sin(A/2)sin(C/2) = sin(B/2) ⇒ 2{√(s-b)(s-c)√bc} × {√(s-a)(s-b)√ab} = √(s-a)(s-c)√ac ⇒ 2(s – b) = b ⇒ a + b + c – 2b = b ⇒ a + c – b = b ⇒ a + c = 2b

Question 4. The value of cos 5π is (a) 0 (b) 1 (c) -1 (d) None of these

Answer: (c) -1 Hint: Given, cos 5π = cos (π + 4π) = cos π = -1

Question 5. In a triangle ABC, cosec A (sin B cos C + cos B sin C) equals (a) none of these (b) c/a (c) 1 (d) a/c

Answer: (c) 1 Hint: Given cosec A (sin B cos C + cos B sin C) = cosec A × sin(B+C) = cosec A × sin(180 – A) = cosec A × sin A = cosec A × 1/cosec A = 1

Question 6. If the angles of a triangle be in the ratio 1 : 4 : 5, then the ratio of the greatest side to the smallest side is (a) 4 : (√5 – 1) (b) 5 : 4 (c) (√5 – 1) : 4 (d) none of these

Answer: (a) 4 : (√5 – 1) Hint: Given, the angles of a triangle be in the ratio 1 : 4 : 5 ⇒ x + 4x + 5x = 180 ⇒ 10x = 180 ⇒ x = 180/10 ⇒ x = 18 So, the angle are: 18, 72, 90 Since a : b : c = sin A : sin B : sin C ⇒ a : b : c = sin 18 : sin 72 : sin 90 ⇒ a : b : c = (√5 – 1)/4 : {√(10 + 2√5)}/4 : 1 ⇒ a : b : c = (√5 – 1) : {√(10 + 2√5)} : 4 Now, c /a = 4/(√5 – 1) ⇒ c : a = 4 : (√5 – 1)

Question 7. The value of cos 180° is (a) 0 (b) 1 (c) -1 (d) infinite

Answer: (c) -1 Hint: 180 is a standard degree generally we all know their values but if we want to go theoretically then cos(90 + x) = – sin(x) So, cos 180 = cos(90 + 90) = -sin 90 = -1 {sin 90 = 1} So, cos 180 = -1

Question 8. The perimeter of a triangle ABC is 6 times the arithmetic mean of the sines of its angles. If the side b is 2, then the angle B is (a) 30° (b) 90° (c) 60° (d) 120°

Answer: (b) 90° Hint: Let the lengths of the sides if ∆ABC be a, b and c Perimeter of the triangle = 2s = a + b + c = 6(sinA + sinB + sinC)/3 ⇒ (sinA + sinB + sinC) = ( a + b + c)/2 ⇒ (sinA + sinB + sinC)/( a + b + c) = 1/2 From sin formula,Using sinA/a = sinB/b = sinC/c = (sinA + sinB + sinC)/(a + b + c) = 1/2 Now, sinB/b = 1/2 Given b = 2 So, sinB/2 = 1/2 ⇒ sinB = 1 ⇒ B = π/2

Question 9: If 3 × tan(x – 15) = tan(x + 15), then the value of x is (a) 30 (b) 45 (c) 60 (d) 90

Answer: (b) 45 Hint: Given, 3×tan(x – 15) = tan(x + 15) ⇒ tan(x + 15)/tan(x – 15) = 3/1 ⇒ {tan(x + 15) + tan(x – 15)}/{tan(x + 15) – tan(x – 15)} = (3 + 1)/(3 – 1) ⇒ {tan(x + 15) + tan(x – 15)}/{tan(x + 15) – tan(x – 15)} = 4/2 ⇒ {tan(x + 15) + tan(x – 15)}/{tan(x + 15) – tan(x – 15)} = 2 ⇒ sin(x + 15 + x – 15)/sin(x + 15 – x + 15) = 2 ⇒ sin 2x/sin 30 = 2 ⇒ sin 2x/(1/2) = 2 ⇒ 2 × sin 2x = 2 ⇒ sin 2x = 1 ⇒ sin 2x = sin 90 ⇒ 2x = 90 ⇒ x = 45

Question 10. If the sides of a triangle are 13, 7, 8 the greatest angle of the triangle is (a) π/3 (b) π/2 (c) 2π/3 (d) 3π/2

Answer: (c) 2π/3 Hint: Given, the sides of a triangle are 13, 7, 8 Since greatest side has greatest angle, Now Cos A = (b² + c² – a²)/2bc ⇒ Cos A = (7² + 8² – 13²)/(2×7×8) ⇒ Cos A = (49 + 64 – 169)/(2×7×8) ⇒ Cos A = (113 – 169)/(2×7×8) ⇒ Cos A = -56/(2×56) ⇒ Cos A = -1/2 ⇒ Cos A = Cos 2π/3 ⇒ A = 2π/3 So, the greatest angle is = 2π/3

Question 11. The value of tan 20 × tan 40 × tan 80 is (a) tan 30 (b) tan 60 (c) 2 tan 30 (d) 2 tan 60

Answer: (b) tan 60 Hint: Given, tan 20 × tan 40 × tan 80 = tan 40 × tan 80 × tan 20 = [{sin 40 × sin 80}/{cos 40 × cos 80}] × (sin 20/cos 20) = [{2 * sin 40 × sin 80}/{2 × cos 40 × cos 80}] × (sin 20/cos 20) = [{cos 40 – cos 120}/{cos 120 + cos 40}] × (sin 20/cos 20) = [{cos 40 – cos (90 + 30)}/{cos (90 + 30) + cos 40}] × (sin 20/cos 20) = [{cos 40 + sin30}/{-sin30 + cos 40}] × (sin 20/cos 20) = [{(2 × cos 40 + 1)/2}/{(-1 + cos 40)/2}] × (sin 20/cos 20) = [{2 × cos 40 + 1}/{-1 + cos 40}] × (sin 20/cos 20) = [{2 × cos 40 × sin 20 + sin 20}/{-cos 20 + cos 40 × cos 20}] = (sin 60 – sin 20 + sin 20)/(-cos 20 + cos 60 + cos 20) = sin 60/cos 60 = tan 60 So, tan 20 × tan 40 × tan 80 = tan 60

Question 12. If the angles of a triangle be in the ratio 1 : 4 : 5, then the ratio of the greatest side to the smallest side is (a) 4 : (√5 – 1) (b) 5 : 4 (c) (√5 – 1) : 4 (d) none of these

Question 13. The general solution of √3 cos x – sin x = 1 is (a) x = n × π + (-1)n × (π/6) (b) x = π/3 – n × π + (-1)n × (π/6) (c) x = π/3 + n × π + (-1)n × (π/6) (d) x = π/3 – n × π + (π/6)

Answer: (c) x = π/3 + n × π + (-1)n × (π/6) Hint: √3 cos x-sin x=1 ⇒ (√3/2)cos x – (1/2)sin x = 1/2 ⇒ sin 60 × cos x – cos 60 × sin x = 1/2 ⇒ sin (x – 60) = 1/2 ⇒ sin (x – π/3) = sin 30 ⇒ sin (x – π/3) = sinπ/6 ⇒ x – π/3 = n × π + (-1)n × (π/6) {where n ∈ Z} ⇒ x = π/3 + n × π + (-1)n × (π/6)

Question 14. If tan² θ = 1 – e², then the value of sec θ + tan³ θ × cosec θ is (a) 2 – e² (b) (2 – e²) 1/2 (c) (2 – e²)² (d) (2 – e²) 3/2

Question 15. The value of cos 20 + 2sin² 55 – √2 sin65 is (a) 0 (b) 1 (c) -1 (d) None of these

Answer: (b) 1 Hint: Given, cos 20 + 2sin² 55 – √2 sin65 = cos 20 + 1 – cos 110 – √2 sin65 {since cos 2x = 1 – 2sin² x} = 1 + cos 20 – cos 110 – √2 sin65 = 1 – 2 × sin {(20 + 110)/2 × sin{(20 – 110)/2} – √2 sin65 {Apply cos C – cos D formula} = 1 – 2 × sin 65 × sin (-45) – √2 sin65 = 1 + 2 × sin 65 × sin 45 – √2 sin65 = 1 + (2 × sin 65)/√2 – √2 sin65 = 1 + √2 ( sin 65 – √2 sin 65 = 1 So, cos 20 + 2sin² 55 – √2 sin65 = 1

Question 16. If the radius of the circumcircle of an isosceles triangle PQR is equal to PQ ( = PR), then the angle P is (a) 2π/3 (b) π/3 (c) π/2 (d) π/6

Answer: (a) 2π/3 Hint: Let S be the center of the circumcircle of triangle PQR. So, SP = SQ = SR = PQ = PR, where SP, SQ & SR are radii. Thus SPQ & SPR are equilateral triangles. ⇒ ∠QSP = 60°; Similarly ∠RQP = 60° ⇒ Angle at the center QSP = 120° So, SRPQ is a rhombus, since all the four sides are equal. Hence, its opposite angles are equal; so ∠P = ∠QSP = 120°

Question 17. If cos a + 2cos b + cos c = 2 then a, b, c are in (a) 2b = a + c (b) b² = a × c (c) a = b = c (d) None of these

Answer: (a) 2b = a + c Hint: Given, cos A + 2 cos B + cos C = 2 ⇒ cos A + cos C = 2(1 – cos B) ⇒ 2 cos((A + C)/2) × cos((A-C)/2 = 4 sin² (B/2) ⇒ 2 sin(B/2)cos((A-C)/2) = 4sin² (B/2) ⇒ cos((A-C)/2) = 2sin (B/2) ⇒ cos((A-C)/2) = 2cos((A+C)/2) ⇒ cos((A-C)/2) – cos((A+C)/2) = cos((A+C)/2) ⇒ 2sin(A/2)sin(C/2) = sin(B/2) ⇒ 2{√(s-b)(s-c)√bc} × {√(s-a)(s-b)√ab} = √(s-a)(s-c)√ac ⇒ 2(s – b) = b ⇒ a + b + c – 2b = b ⇒ a + c – b = b ⇒ a + c = 2b

Question 18. The value of 4 × sin x × sin(x + π/3) × sin(x + 2π/3) is (a) sin x (b) sin 2x (c) sin 3x (d) sin 4x

Answer: (c) sin 3x Hint: Given, 4 × sin x × sin(x + π/3) × sin(x + 2π/3) = 4 × sin x × {sin x × cos π/3 + cos x × sin π/3} × {sin x × cos 2π/3 + cos x × sin 2π/3} = 4 × sin x × {(sin x)/2 + (√3 × cos x)/2} × {-(sin x)/2 + (√3 × cos x)/2} = 4 × sin x × {-(sin 2x)/4 + (3 × cos 2x)/4} = sin x × {-sin 2x + 3 × cos 2x} = sin x × {-sin 2x + 3 × (1 – sin 2x)} = sin x × {-sin 2x + 3 – 3 × sin 2x} = sin x × {3 – 4 × sin 2x} = 3 × sin x – 4 sin 3x = sin 3x So, 4 × sin x × sin(x + π/3) × sin(x + 2π/3) = sin 3x

Question 19. If tan A – tan B = x and cot B – cot A = y, then the value of cot (A – B) is (a) x + y (b) 1/x + y (c) x + 1/y (d) 1/x + 1/y

Answer: (d) 1/x + 1/y Hint: Given, tan A – tan B = x ……………. 1 and cot B – cot A = y ……………. 2 From equation, 1/cot A – 1/cot B = x ⇒ (cot B – cot A)/(cot A × cot B) = x ⇒ y/(cot A × cot B) = x {from equation 2} ⇒ y = x × (cot A × cot B) ⇒ cot A × cot B = y/x Now, cot (A – B) = (cot A × cot B + 1)/(cot B – cot A) ⇒ cot (A – B) = (y/x + 1)/y ⇒ cot (A – B) = (y/x) × (1/y) + 1/y ⇒ cot (A – B) = 1/x + 1/y

Question 20. The value of (sin 7x + sin 5x) /(cos 7x + cos 5x) + (sin 9x + sin 3x) / (cos 9x + cos 3x) is (a) tan 6x (b) 2 tan 6x (c) 3 tan 6x (d) 4 tan 6x

Answer: (b) 2 tan 6x Hint: Given, (sin 7x + sin 5x) /(cos 7x + cos 5x) + (sin 9x + sin 3x) / (cos 9x + cos 3x) ⇒ [{2 × sin(7x+5x)/2 × cos(7x-5x)/2}/{2 × cos(7x+5x)/2 × cos(7x-5x)/2}] + [{2 × sin(9x+3x)/2 × cos(9x-3x)/2}/{2 × cos(9x+3x)/2 × cos(9x-3x)/2}] ⇒ [{2 × sin 6x × cosx}/{2 × cos 6x × cosx}] + [{2 × sin 6x × cosx}/{2 × cos 6x × cosx}] ⇒ (sin 6x/cos 6x) + (sin 6x/cos 6x) ⇒ tan 6x + tan 6x ⇒ 2 tan 6x

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Chapter 3 Class 11 Trigonometric Functions

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NCERT Solutions of Chapter 3 Class 11 Trigonometry is available free at teachoo. You can check the detailed explanation of all questions of exercises, examples and miscellaneous by clicking on the Exercise link below.

We had learned Basics of Trigonometry in Class 10. In this chapter, we will learn

• What is a positive or a negative angle
• Measuring angles in Degree , Minutes and Seconds
• Radian measure of an angle
• Converting Degree to Radians , and vice-versa
• Sign of sin, cos, tan in all 4 quadrants
• Finding values of trigonometric functions when one value is given (Example: Finding value of sin, cot, cosec, tan, sec, when cos x = -3/5 is given)
• Finding Value of trigonometric functions, given angle
• Solving questions by formula like  (x + y) formula, 2x 3x formula, Cos x + cos y formula , 2 sin x sin y formula 
• Finding principal and general solutions of a trigonometric equation
• Sin and Cosine Formula with supplementary Questions

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• Important Questions for CBSE Class 11 Maths Chapter 3 Trigonometric Functions 2023-24

CBSE Class 11 Maths Chapter-3 Important Questions - Free PDF Download

Trigonometric functions class 11 important questions have been prepared for students of class 11 to help them score better marks in the examination. The complete topic of trigonometric functions is designed by the subject experts following the latest guidelines of CBSE . The class 11th maths trigonometric functions important questions feature the step by step solutions for easy to difficult questions according to the understanding level of the students. Moreover, students can download the PDF study materials from Vedantu that will help them to take a ready reference of the chapters and questions whenever they need. Based on the important questions, students can also make vital subject notes and mark them for a quick revision. As the students solve these important questions, they will easily develop a command over the topic of trigonometric functions. 

Download CBSE Class 11 Maths Important Questions 2023-24 PDF

Also, check CBSE Class 11 Maths Important Questions for other chapters:

Study Important Questions for Class 11 Mathematics Chapter 3 – Trigonometric Functions

1 Mark Questions

1. Find the radian measure corresponding to  $ 5{}^\circ \text{ }37'\text{ }30'' $  

Ans-  

Converting the given value to a pure degree form

 $ {{5}^{\circ }}37'30''={{5}^{\circ }}37'\left( \dfrac{30}{60} \right)' $ 

 $ \Rightarrow {{5}^{\circ }}37'60''={{5}^{\circ }}\left( \dfrac{75}{2} \right)' $ 

 $ \Rightarrow {{5}^{\circ }}37'60''={{5}^{\circ }}{{\left( \dfrac{75}{2\left( 60 \right)} \right)}^{\circ }} $ 

 $ \Rightarrow {{5}^{\circ }}37'60''={{\left( \dfrac{45}{8} \right)}^{\circ }} $ 

Degree to Radian Conversion

 $ \left( \dfrac{45}{8} \right)\left( \dfrac{\pi }{180} \right)=\dfrac{\pi }{32}\text{rad} $ 

2. Find degree measure corresponding to  $ {{\left( \dfrac{\pi }{16} \right)}^{c}} $  

Converting the given value from radian to degree form

$ \dfrac{\pi }{16}\times \dfrac{180}{\pi }={{\left( \dfrac{45}{4} \right)}^{\circ }} $ 

Simplify degree form

$ {{\left( \dfrac{45}{4} \right)}^{\circ }}={{11}^{\circ }}15' $ 

3. Find the length of an arc of a circle of radius  $ 5cm $  subtending a central angle measuring  $ 15{}^\circ  $  

The arc of a circle with a radius of  $ 5\,\text{cm} $  with a central angle of  $ {{15}^{\circ }} $  should be of the length  $ \dfrac{5\pi }{12}cm $  using the formula  $ \text{Arc}\,\text{=}\,\pi \times \left( \theta  \right) $ .

4. Find the value of  $ \dfrac{19\pi }{3} $  

We have  $ \tan \dfrac{19\pi }{3} $ 

 $ \tan \dfrac{19\pi }{3}=\tan \left( 6\dfrac{\pi }{3} \right) $ 

                   $ =\tan \left( 6\pi +\dfrac{\pi }{3} \right) $ 

                   $ =\tan \left( 3\times 2\pi +\dfrac{\pi }{3} \right) $ 

                   $ =\tan \left( \dfrac{\pi }{3} \right) $ 

                   $ =\sqrt{3} $ 

5. Find the value of  $ \sin \left( -1125{}^\circ  \right) $  

We have  $ \sin \left( -{{1125}^{\circ }} \right) $ 

 $ \sin \left( -\dfrac{1125}{360}\times {{360}^{\circ }} \right) $ 

 $ =-\sin \left( \left( 3+\dfrac{45}{360} \right)\times {{360}^{\circ }} \right) $ 

 $ =-\sin \left( {{45}^{\circ }} \right) $ 

 $ =-\dfrac{1}{\sqrt{2}} $ 

6. Find the value of  $ \tan \left( {{15}^{\circ }} \right) $  

We have  $ \tan {{15}^{\circ }} $ 

 $ \tan {{15}^{\circ }}=\tan \left( {{60}^{\circ }}-{{45}^{\circ }} \right) $ 

                 $ =\dfrac{\tan {{60}^{\circ }}-\tan {{45}^{\circ }}}{1+\tan {{60}^{\circ }}\times \tan {{45}^{\circ }}} $ 

                 $ =\dfrac{\sqrt{3}-1}{\sqrt{3}+1} $ 

7. If  $ \sin A=\dfrac{3}{5} $   and  $ \dfrac{\pi }{2}<A< $  find $ \cos A $  

The condition  $ \dfrac{\pi }{2}<A $  denotes that we need to take into account for the second quadrant , hence the cosine value will be negative . 

Therefore, 

 $ \cos A=\dfrac{-4}{5} $ 

8. If  $ \tan A=\dfrac{a}{a+1} $   and  $ \tan B=\dfrac{1}{2a+1} $   then find the value of  $ A+B $  

 $ \tan \left( A+B \right)=\dfrac{\tan A+\tan B}{1-\tan A\tan B} $ 

                           $ =\dfrac{\dfrac{a}{a+1}+\dfrac{1}{2a+1}}{1-\dfrac{a}{a+1}\cdot \dfrac{1}{2a+1}} $ 

                          $ =\dfrac{\dfrac{2{{a}^{2}}+2a+1}{\left( a+1 \right)\left( 2a+1 \right)}}{\dfrac{\left( a+1 \right)\left( 2a+1 \right)-a}{\left( a+1 \right)\left( 2a+1 \right)}} $ 

                          $ =1 $ 

Which can only be possible if  $ A+B={{45}^{\circ }} $ .

9. Express  $ \sin 12\theta +\sin 4\theta  $   as the product of sines and cosine

Using the trigonometric difference formula, we get

 $ \sin 12\theta +\sin 4\theta =\sin \left( 8\theta +4\theta  \right)+\sin \left( 8\theta -4\theta  \right) $  

                                    $ =2\sin 8\theta \cos 4\theta  $ 

10. Express  $ 2\cos 4x\sin 2x $   as an algebraic sum of sines or cosine.

 $ 2\cos 4x\sin 2x=\sin \left( 2x+4x \right)+\sin \left( 2x-4x \right) $ 

                                 $ =\sin 6x+\sin \left( -2x \right) $ 

                                  $ =\sin 6x-\sin 2x $ 

11. Write the range of  $ \cos \theta  $  

The cosine function is a periodic function with a domain of  $ \mathbb{R} $  and a range of  $ \left[ -1,1 \right] $ .

12. What is domain of  $ \sec \theta  $  

The secant function is the reciprocal of the cosine function , it has a domain of  $ \mathbb{R}-\left\{ (2n+1)\dfrac{\pi }{2};n\in \mathbb{Z} \right\} $  because those are the points where the cosine function equates to  $ 0 $ .

13. Find the principal solution of  $ \cot x=3 $  

The principal solution of  $ \cot x=3 $  is for the following input values  $ x=\dfrac{5\pi }{6},\dfrac{11\pi }{6} $ .

14. Write the general solution of  $ \cos \theta =0 $  

The general solution for the equation  $ \cos \theta =0 $  is  $ \theta =(2n+1)\dfrac{\pi }{2},n\in \mathbb{Z} $ .

15. If  $ \sin x=\dfrac{\sqrt{5}}{3} $ and  $ 0\text{ }<\text{ }x\text{ }<\dfrac{\pi }{2} $ find the value of $ \cos 2x $  

We know that  $ \cos 2x=1-{{\sin }^{2}}x $ 

 $ \cos 2x=1-2{{\left( \dfrac{\sqrt{5}}{3} \right)}^{2}} $ 

                $ =1-2\times \dfrac{5}{9} $ 

                $ =-\dfrac{1}{9} $ 

16. If  $ \cos x=-\dfrac{1}{3} $   and  $ x $   lies in quadrant  $ \text{III} $ , find the value of  $ \sin \dfrac{x}{2} $  

We know that  $ \cos 2x=1-2{{\sin }^{2}}x $ 

 $ \cos \left( 2\left( \dfrac{x}{2} \right) \right)=1-2{{\sin }^{2}}\left( \dfrac{x}{2} \right) $ 

 $ \Rightarrow -\dfrac{1}{3}=1-2{{\sin }^{2}}\dfrac{x}{2} $ 

 $ \Rightarrow 2{{\sin }^{2}}\dfrac{x}{2}=1+\dfrac{1}{3} $ 

 $ \Rightarrow {{\sin }^{2}}\dfrac{x}{2}=\dfrac{2}{3} $ 

 $ \Rightarrow \sin \dfrac{x}{2}=\pm \sqrt{\dfrac{2}{3}} $ 

 $ \Rightarrow \sin \dfrac{x}{2}=\sqrt{\dfrac{2}{3}}\,\,\,\,\,\left[ \text{2nd Quadrant} \right] $ 

17. Convert into radian measures  $ -47{}^\circ 30' $  

Convert into pure degree form and then convert to radian

 $ -47{}^\circ 30'=-{{\left( 47+\dfrac{30}{60} \right)}^{{}^\circ }} $ 

                   $ =-{{\left( 47+\dfrac{1}{2} \right)}^{{}^\circ }} $ 

                   $ =-\left( \dfrac{95}{2}\times \dfrac{\pi }{180} \right)\text{rad} $ 

                   $ =-\dfrac{19\pi }{72}\text{rad} $ 

18. Evaluate  $ \tan 75{}^\circ  $  

Use the trigonometric addition formula for the tangent function

 $ \tan {{75}^{\circ }}=\tan ({{45}^{\circ }}+{{30}^{\circ }}) $ 

                 $ =\dfrac{\tan {{45}^{\circ }}+\tan {{30}^{\circ }}}{1-\tan {{45}^{\circ }}\tan {{30}^{\circ }}} $ 

                 $ =\dfrac{\sqrt{3}+1}{\sqrt{3}-1} $ 

19. Prove that  $ \sin (40+\theta )\cdot \cos (10+\theta )-\cos (40+\theta )\cdot \sin (10+\theta )=\dfrac{1}{2} $  

Ans-  

Let us take the left-hand side of the equation and make some manipulations.

We know,  $ \sin \left( a-b \right)=\sin a\cos b-\cos a\sin b $ 

 $ \text{L}\text{.H}\text{.S}=\sin (40+\theta )\cos (10+\theta )-\cos (40+\theta )\sin (10+\theta ) $ 

              $ =\sin \left[ 40+\theta -10-\theta  \right]=\sin 30 $ 

              $ =\dfrac{1}{2} $ 

20. Find the principal solution of the eq.  $ \sin x=\dfrac{\sqrt{3}}{2} $  

The principal solution of  $ \sin x=\dfrac{\sqrt{3}}{2} $  is the input values of  $ x=\dfrac{\pi }{3},\dfrac{2\pi }{3} $ 

21. Prove that  $ \cos \left( \dfrac{\pi }{4}+x \right)+\cos \left( \dfrac{\pi }{4}-x \right)=\sqrt{2}\cos x $ 

Let us start with the left-hand side and use the trigonometric differences formula for the cosine function

 $ \text{L}\text{.H}\text{.S}=\cos \left( \dfrac{\pi }{4}+x \right)+\cos \left( \dfrac{\pi }{4}-x \right) $ 

               $ =2\cos \dfrac{\pi }{4}\cos x $ 

               $ =2\left( \dfrac{1}{\sqrt{2}} \right)\cos x $ 

               $ =\sqrt{2}\cos x $ 

               $ =\text{R}\text{.H}\text{.S} $ 

22. Convert into radian measures  $ -37{}^\circ 30' $  

Convert into pure degree form and then convert from degree to radian

 $ -37{}^\circ 30'={{\left( 37+\dfrac{30}{60} \right)}^{{}^\circ }} $ 

                 $ =-{{\left( \dfrac{75}{2} \right)}^{{}^\circ }} $ 

                 $ =-\dfrac{75}{2}\times \dfrac{\pi }{180}\text{rad} $ 

                 $ =-\dfrac{5\pi }{24}\text{rad} $ 

 $ Sin\text{ }\left( n+1 \right)\text{ }x\text{ }Sin\text{ }\left( n+2 \right)\text{ }x\text{ }+\text{ }Cos\text{ }\left( n+1 \right)\text{ }x.\text{ }Cos\text{ }\left( n+2 \right)\text{ }x\text{ }=\text{ }Cos\text{ }x $   

 $ \text{L}\text{.H}\text{.S}\,\text{. = sin}\left( n+1 \right)x\sin \left( n+2 \right)x+\cos \left( n+1 \right)x\cos \left( n+2 \right)x $ 

                 $ =\cos \left\{ \left( n+1 \right)x-\left( n+2 \right)x \right\} $ 

                 $ =\cos \left( nx+x-n-2x \right) $ 

                 $ =\cos \left( -x \right) $ 

                 $ =\cos \left( x \right) $ 

Find the value of  $ \operatorname{Sin}\dfrac{31\pi }{3} $  

We have  $ \sin \dfrac{31\pi }{3} $ 

 $ \operatorname{Sin}\dfrac{31\pi }{3}=\operatorname{Sin}\left( 10\pi +\dfrac{\pi }{3} \right) $ 

                   $ =\operatorname{Sin}\left( 2\pi \times 5+\dfrac{\pi }{3} \right)\,\,\,\,\,\,\left[ \text{Periodic Function} \right] $ 

                   $ =\operatorname{Sin}\dfrac{\pi }{3} $ 

                   $ =\dfrac{\sqrt{3}}{2} $ 

Find the principal solution of the eq.  $ \tan x=-\dfrac{1}{\sqrt{3}} $ . 

The principal solution of the equation  $ \tan x=-\dfrac{1}{\sqrt{3}} $  will be the input values of  $ x=\dfrac{5\pi }{6},\dfrac{11\pi }{6} $ 

Convert into radian measures $ 5{}^\circ \text{ }37'\text{ }30'' $  

  Prove  $ Cos70{}^\circ .\text{ }Cos10{}^\circ +\text{ }Sin70{}^\circ .\text{ }Sin10{}^\circ =\dfrac{1}{2} $  

Starting with the left-hand side and using the trigonometric differences formula for the cosine function .

 $ \text{L}\text{.H}\text{.S}=\text{cos}\left( {{70}^{\circ }}{{10}^{\circ }} \right) $ 

               $ =\cos {{60}^{\circ }} $ 

               $ =\dfrac{1}{2} $ 

  Evaluate   $ 2\operatorname{Sin}\dfrac{\pi }{12} $  

Use the trigonometric difference formula for the sine function and expand

 $ 2\sin \dfrac{\pi }{12}=2\sin \left[ \dfrac{\pi }{4}-\dfrac{\pi }{6} \right] $ 

                   $ =2\left[ \sin \dfrac{\pi }{4}\cos \dfrac{\pi }{6}-\cos \dfrac{\pi }{4}\sin \dfrac{\pi }{6} \right] $ 

                   $ =2\left[ \dfrac{1}{\sqrt{2}}\times \dfrac{\sqrt{3}}{2}-\dfrac{1}{\sqrt{2}}\times \dfrac{1}{2} \right] $ 

                    $ =\dfrac{\sqrt{3}-1}{\sqrt{2}} $ 

 Find the solution of  $ \operatorname{Sin}x=-\dfrac{\sqrt{3}}{2} $  

We are required to find the general solution for the equation  $ \sin x=-\dfrac{\sqrt{3}}{2} $ 

 $ \operatorname{Sin}x=-\dfrac{\sqrt{3}}{2} $ 

 $ \Rightarrow \operatorname{Sin}x=\operatorname{Sin}\left( \pi +\dfrac{\pi }{3} \right) $ 

 $ \Rightarrow \operatorname{Sin}x=\operatorname{Sin}\dfrac{4\pi }{3} $ 

 $ \operatorname{Sin}\theta =\operatorname{Sin}\alpha  $ 

 $ \theta =n\pi +{{(-1)}^{n}}\cdot \alpha  $ 

 $ x=n\pi +{{(-1)}^{n}}\cdot \dfrac{4\pi }{3} $ 

 Prove that  $ \dfrac{\operatorname{Cos}9{}^\circ -\operatorname{Sin}9{}^\circ }{\operatorname{Cos}9{}^\circ +\operatorname{Sin}9{}^\circ }=\tan 36{}^\circ  $  

Let us start with the right-hand side and use the trigonometric differences formula for the tangent function.

 $ \text{R}\text{.H}\text{.S}=\tan 36{}^\circ  $ 

                $ =\tan \left( {{45}^{\circ }}-{{9}^{\circ }} \right) $ 

                $ =\dfrac{\tan {{45}^{\circ }}-\tan {{9}^{\circ }}}{1+\tan {{45}^{\circ }}\tan {{9}^{\circ }}} $ 

                $ =\dfrac{1-\tan {{9}^{\circ }}}{1+\tan {{9}^{\circ }}} $ 

                $ =\dfrac{\cos {{9}^{\circ }}-\sin {{9}^{\circ }}}{\cos {{9}^{\circ }}+\sin {{9}^{\circ }}} $ 

                $ =\text{L}\text{.H}\text{.S}\text{.} $ 

 Find the value of  $ \tan \dfrac{19\pi }{3} $  

We have  $ \tan \left( \dfrac{19\pi }{3} \right) $ 

 $ \tan \dfrac{19\pi }{3}=\tan \left( 6\pi -\dfrac{\pi }{3} \right) $ 

                  $ =\tan \left[ 3\times 2\pi +\dfrac{\pi }{3} \right]\,\,\,\,\,\,\,\,\,\,\left[ \text{Periodic Function} \right] $ 

                  $ =\tan \dfrac{\pi }{3} $ 

                  $ =\sqrt{3} $ 

  Prove   $ \operatorname{Cos}4x=1-8{{\operatorname{Sin}}^{2}}x.{{\operatorname{Cos}}^{2}}x $ 

Starting with the left-hand side and using the trigonometric addition formula, $ \cos 2x=1-2{{\sin }^{2}}x $ 

 $ \text{L}\text{.H}\text{.S}=\operatorname{Cos}4x $ 

               $ =1-2{{\operatorname{Sin}}^{2}}2x $ 

               $ =1-2{{(\operatorname{Sin}2x)}^{2}} $ 

               $ =1-2{{(2\operatorname{Sin}x.\operatorname{Cos}x)}^{2}} $ 

               $ =1-2(4{{\operatorname{Sin}}^{2}}x.{{\operatorname{Cos}}^{2}}x) $ 

               $ =1-8{{\operatorname{Sin}}^{2}}x.{{\operatorname{Cos}}^{2}}x $ 

 Prove  $ \dfrac{\operatorname{Cos}(\pi +x).\operatorname{Cos}(-x)}{\operatorname{Sin}(\pi -x).\operatorname{Cos}\left( \dfrac{\pi }{2}+x \right)}=Co{{t}^{2}}x $  

Starting with the left-hand side and using the trigonometric periodic identities, we obtain the following

 $ \text{L}\text{.H}\text{.S}\text{.}=\dfrac{\cos \left( \pi +x \right)\cos \left( -x \right)}{\sin \left( \pi -x \right)\cos \left( \dfrac{\pi }{2}+x \right)} $ 

                 $ =\dfrac{-\cos x\cos x}{-\sin x\sin x} $ 

                 $ ={{\cot }^{2}}x $ 

                 $ =\text{R}\text{.H}\text{.S}\text{.} $ 

 Prove that  $ \tan {{56}^{\circ }}=\dfrac{\operatorname{Cos}{{11}^{\circ }}+\operatorname{Sin}{{11}^{\circ }}}{\operatorname{Cos}{{11}^{\circ }}-\operatorname{Sin}{{11}^{\circ }}} $  

Starting with the left-hand side and using the trigonometric addition formula for the tangent function, we obtain

 $ \text{L}\text{.H}\text{.S}\text{.}=\tan {{56}^{\circ }} $ 

                 $ =\tan ({{45}^{\circ }}+{{11}^{\circ }}) $ 

                 $ =\dfrac{\tan {{45}^{\circ }}+\tan {{11}^{\circ }}}{1-\tan {{45}^{\circ }}\cdot \tan {{11}^{\circ }}} $ 

                 $ =\dfrac{1+\tan {{11}^{\circ }}}{1-\tan {{11}^{\circ }}} $ 

                $ =\dfrac{\cos {{11}^{\circ }}+\sin {{11}^{\circ }}}{\cos {{11}^{\circ }}-\sin {{11}^{\circ }}} $ 

                $ =\text{R}\text{.H}\text{.S}\text{.} $ 

  Prove that  $ \operatorname{Cos}{{105}^{\circ }}+\operatorname{Cos}{{15}^{\circ }}=\operatorname{Sin}{{75}^{\circ }}-\operatorname{Sin}{{15}^{\circ }} $ 

Starting with the left-hand side and using the trigonometric difference formula for the cosine function, we obtain

 $ \text{L}\text{.H}\text{.S}\text{.}=\operatorname{Cos}{{105}^{\circ }}+\operatorname{Cos}{{15}^{\circ }} $ 

                 $ =\operatorname{Cos}({{90}^{\circ }}+{{15}^{\circ }})+\operatorname{Cos}({{90}^{\circ }}-{{75}^{\circ }}) $ 

                 $ =-\operatorname{Sin}{{15}^{\circ }}+\operatorname{Sin}{{75}^{\circ }} $ 

                 $ =\operatorname{Sin}{{75}^{\circ }}-\operatorname{Sin}{{15}^{\circ }} $ 

  Find the value of   $ \operatorname{Cos}(-{{1710}^{\circ }}) $ 

We have  $ \cos \left( -{{1710}^{\circ }} \right) $ . We also know  $ \cos \left( -x \right)=\cos x $ 

 $ \operatorname{Cos}(-{{1710}^{\circ }})=\operatorname{Cos}(1800-90) $ 

                               $ =\operatorname{Cos}\left[ 5\times 360+90 \right] $ 

                               $ =\operatorname{Cos}\dfrac{\pi }{2} $ 

                               $ =0 $ 

   A wheel makes  $ 360 $  revolutions in  $ 1 $  minute. Through how many radians does it turn in  $ 1 $  second.

 $ \text{Number of revolutions made in 60s}=360 $ 

 $ \text{Number of revolutions made in 1s}=\dfrac{360}{60} $ 

 $ \text{Angle moved in 6 revolutions}=2\pi \times 6 $ 

 $ =12\pi  $ 

Prove that  $ {{\operatorname{Sin}}^{2}}6x-{{\operatorname{Sin}}^{2}}4x=\operatorname{Sin}2x.\operatorname{Sin}10x $ 

Starting with the left-hand side and using the trigonometric addition formula for the sine function, we obtain

 $ \text{L}\text{.H}\text{.S}\text{.}={{\operatorname{Sin}}^{2}}6x-{{\operatorname{Sin}}^{2}}4x $ 

                $ =\sin \left( 6x+4x \right)\sin \left( 6x-4x \right) $ 

                $ =\sin 10x\sin 2x $ 

Prove that  $ \dfrac{\tan 69+\tan 66}{1\tan 69.\tan 66}=-1 $ 

Starting with the left-hand side and using the trigonometric difference identity for the tangent function, we obtain

 $ \text{L}\text{.H}\text{.S}\text{.}=\dfrac{\tan {{69}^{\circ }}+\tan {{66}^{\circ }}}{1-\tan {{69}^{\circ }}\tan {{66}^{\circ }}} $ 

                 $ =\tan ({{69}^{\circ }}+{{66}^{\circ }}) $ 

                 $ =\tan \left( {{135}^{\circ }} \right) $ 

                 $ =\tan \left( {{90}^{\circ }}+{{45}^{\circ }} \right) $ 

                $ =-1 $ 

 Prove that  $\dfrac{\operatorname{Sin}x}{1+\operatorname{Cos}x}=\tan 

\dfrac{x}{2} $  

Starting with the left-hand side and using the trigonometric addition identities for the sine and cosine function, we obtain

 $ \text{L}\text{.H}\text{.S}\text{.}=\dfrac{\sin x}{1+\cos x} $ 

                 $ =\dfrac{2\sin \dfrac{x}{2}\cos \dfrac{x}{2}}{2{{\cos }^{2}}\dfrac{x}{2}} $ 

                 $ =\dfrac{\sin \dfrac{x}{2}}{\cos \dfrac{x}{2}} $ 

                 $ =\tan \dfrac{x}{2} $ 

4 Marks Questions

Prove the following identities

1. The minute hand of a watch is  $ 1.5cm $   long. How far does its tip move in  $ 40 $  minute? 

Analysing the given information, we have 

 $ r=1.5cm $ 

 $ \text{Angle made in }60\min ={{360}^{\circ }} $ 

 $ \text{Angle made in 1min}={{6}^{\circ }} $ 

 $ \text{Angle made in 40min}={{6}^{\circ }}\times {{40}^{\circ }}={{240}^{\circ }} $ 

Calculating the arc distance

              $ \theta =\dfrac{l}{r} $ 

 $ 240\times \dfrac{\pi }{180}=\dfrac{l}{1.5} $ 

      $ 2\times 3.14=l $ 

              $ 6.28=l $ 

                      $ l=6.28cm $ 

2. Show that  $ tan\text{ }3x.\text{ }tan\text{ }2x.\text{ }tan\text{ }x\text{ }=\text{ }tan\text{ }3x\text{ }\text{ }tan\text{ }2x\text{ }\text{ }tan\text{ }x $  

Let us start with  $ \tan 3x $  and we know  $ 3x=2x+x $ 

 $ \tan 3x=\tan (2x+x) $ 

 $ \dfrac{\tan 3x}{1}=\dfrac{\tan 2x+\tan x}{1-\tan 2x.\tan x} $ 

 $ \tan 3x(1-\tan 2x.\tan x)=\tan 2x+\tan x $ 

 $ \tan 3x-\tan 3x.\tan 2x.\tan x=\tan 2x+\tan x $ 

 $ \tan 3x.\tan 2x.\tan x=\tan 3x-\tan 2x-\tan x $ 

3. Find the value of  $ \tan \dfrac{\pi }{8} $  

We know that

 $ \tan 2x=\dfrac{2\tan x}{1-{{\tan }^{2}}x} $ 

Therefore, we have

    $ \tan \left( 2\dfrac{\pi }{8} \right)=\dfrac{2\tan \dfrac{\pi }{8}}{1-{{\tan }^{2}}\dfrac{\pi }{8}} $ 

 $ \Rightarrow 1=\dfrac{2\tan \dfrac{\pi }{8}}{1-{{\tan }^{2}}\dfrac{\pi }{8}} $ 

Put  $ \tan \dfrac{\pi }{8}=x $ 

 $ 1=\dfrac{2x}{1-{{x}^{2}}} $ 

 $ \Rightarrow 2x=1-{{x}^{2}} $ 

 $ \Rightarrow x=\dfrac{-1\pm \sqrt{2}}{1} $ 

Since,  $ \dfrac{\pi }{8} $  lies in the first quadrant, the value must be positive, hence

$ \tan \dfrac{\pi }{8}=\sqrt{2}-1 $ 

4. Prove that  $ \dfrac{\operatorname{Sin}(x+y)}{\operatorname{Sin}(x-y)}=\dfrac{\tan x+\tan y}{\tan x-\tan y} $  

Starting with the left-hand side and using the trigonometric difference formula for the sine function, we get

 $ \text{L}\text{.H}\text{.S}\text{.}=\dfrac{\operatorname{Sin}(x+y)}{\operatorname{Sin}(x-y)} $ 

                $ =\dfrac{\operatorname{Sin}x.\operatorname{Cos}y+\operatorname{Cos}x.\operatorname{Sin}y}{\operatorname{Sin}x.\operatorname{Cos}y-\operatorname{Cos}x.\operatorname{Sin}y} $ 

Dividing numerator and denominator by  $ \operatorname{Cos}x.\operatorname{Cos}y $ 

           $ =\dfrac{\tan x+\tan y}{\tan x-\tan y} $ 

               $ =\text{R}\text{.H}\text{.S}\text{.} $ 

5. If in two circles, arcs of the same length subtend angles  $ {{60}^{\circ }} $  and  $ {{75}^{\circ }} $  at the center find the ratio of their radii. 

We know that the length of the arc and its subtended angle is related using the following formula

 $ \theta =\dfrac{1}{{{r}_{1}}} $ 

 $ 60\times \dfrac{\pi }{18}=\dfrac{1}{{{r}_{1}}} $ 

 $ {{r}_{1}}=\dfrac{3l}{\pi } $      ….. $ (1) $ 

 $ \theta =\dfrac{1}{{{r}_{2}}} $ 

 $ 75\times \dfrac{\pi }{18}=\dfrac{1}{{{r}_{2}}} $ 

 $ {{r}_{2}}=\dfrac{12l}{5\pi } $     ….. $ (2) $ 

 $ (1)\div (2) $ 

 $ \dfrac{{{r}_{1}}}{{{r}_{2}}}=\dfrac{\dfrac{3l}{\pi }}{\dfrac{12l}{5\pi }} $ 

 $ =\dfrac{31}{\pi }\times \dfrac{5\pi }{12l} $ 

 $ =5:4 $ 

6. Prove that  $ \operatorname{Cos}6x=32{{\operatorname{Cos}}^{2}}x-48{{\operatorname{Cos}}^{4}}x+18{{\operatorname{Cos}}^{2}}x-1 $  

Starting with the left-hand side and using the trigonometric identities for the cosine function, we obtain

 $ \text{L}\text{.H}\text{.S}\text{.}=\operatorname{Cos}6x $ 

                 $ =\operatorname{Cos}2(3x)=2{{\operatorname{Cos}}^{2}}3x-1 $ 

                 $ =\operatorname{Cos}2(3x) $ 

                 $ =2{{(4co{{s}^{3}}x-3\cos x)}^{2}}-1 $ 

                 $ =2\left[ 16{{\operatorname{Cos}}^{6}}x+9{{\operatorname{Cos}}^{2}}x-24{{\operatorname{Cos}}^{4}}x \right]-1 $ 

                  $ =32{{\operatorname{Cos}}^{6}}x+18{{\operatorname{Cos}}^{2}}x-48{{\operatorname{Cos}}^{4}}x-1 $ 

               $ =32{{\operatorname{Cos}}^{6}}x-48{{\operatorname{Cos}}^{4}}x+18{{\operatorname{Cos}}^{2}}x1 $ 

                  $ =\text{R}\text{.H}\text{.S}\text{.} $ 

7. Solve  $ \operatorname{Sin}2x-\operatorname{Sin}4x+\operatorname{Sin}6x=0 $  

Starting with the left-hand side and using the trigonometric addition identity for the sine function, we obtain

 $ \text{L}\text{.H}\text{.S}\text{.}=\operatorname{Sin}6x+\operatorname{Sin}2x-\operatorname{Sin}4x $ 

                 $ =2\sin \left( \dfrac{6x+2x}{2} \right)\cos \left( \dfrac{6x-2x}{2} \right)-\sin 4x $ 

                 $ =\sin 4x\left( 2\cos 2x-1 \right) $ 

                 $ =0 $ 

 $ \sin 4x=0 $ 

      $ 4x=n\pi  $ 

        $ x=\dfrac{n\pi }{4} $ 

 $ 2\cos 2x-1=0 $ 

        $ \cos 2x=\cos \dfrac{\pi }{3} $ 

             $ 2x=2n\pi \pm \dfrac{\pi }{3} $ 

              $ x=n\pi \pm \dfrac{\pi }{6} $ 

8. In a circle of diameter  $ 40cm $ , the length of a chord is  $ 20cm $ . Find the length of the minor area of the chord. 

 $ \theta =\dfrac{l}{r} $ 

 $ \Rightarrow 60\times \dfrac{\pi }{180}=\dfrac{l}{20} $ 

 $ \Rightarrow l=\dfrac{20\pi }{3}\text{cm/s} $ 

9. Prove that  $ \tan 4x=\dfrac{4\tan x(1-{{\tan }^{2}}x)}{1-6{{\tan }^{2}}x+{{\tan }^{4}}x} $  

Starting with the left-hand side and using the trigonometric addition identities for the tangent function, we obtain

 $ \text{L}\text{.H}\text{.S}\text{.}=\tan 4x $ 

                 $ =\dfrac{2\tan 2x}{1-{{\tan }^{2}}2x} $ 

                 $ =\dfrac{2.\dfrac{2\tan 2x}{1-{{\tan }^{2}}2x}}{1-{{\left( \dfrac{2\tan 2x}{1-{{\tan }^{2}}2x} \right)}^{2}}} $ 

                $ =\dfrac{\dfrac{4\tan x}{1-{{\tan }^{2}}x}}{\dfrac{{{(1-{{\tan }^{2}}x)}^{2}}-4{{\tan }^{2}}x}{{{(1-{{\tan }^{2}}x)}^{2}}}} $ 

                 $ =\dfrac{4\tan x}{(1-{{\tan }^{2}}x)}\times \dfrac{(1-{{\tan }^{2}}x)}{1+{{\tan }^{4}}x-2{{\tan }^{2}}x-4{{\tan }^{2}}x} $ 

                 $ =\dfrac{4\tan x(1-{{\tan }^{2}}x)}{1-6{{\tan }^{2}}x+{{\tan }^{4}}x} $ 

10. Prove that  $ {{\left( Cosx+Cosy \right)}^{2}}+{{\left( SinxSiny \right)}^{2}}=4Co{{s}^{2}}\left( \dfrac{x+y}{2} \right) $  

Starting with the left-hand side and using the trigonometric addition identities for the cosine and sine function, we obtain

 $ \text{L}\text{.H}\text{.S}\text{.}={{\left( Cosx+Cosy \right)}^{2}}+{{\left( SinxSiny \right)}^{2}} $ 

                $ ={{\left( 2\operatorname{Cos}\dfrac{x+y}{2}.\operatorname{Cos}\dfrac{x-y}{2} \right)}^{2}}+{{\left( 2\operatorname{Cos}\left( \dfrac{x+y}{2} \right).\operatorname{Sin}\left( \dfrac{x-y}{2} \right) \right)}^{2}} $ 

                $ =4{{\operatorname{Cos}}^{2}}\dfrac{x+y}{2}.{{\operatorname{Cos}}^{2}}\left( \dfrac{x-y}{2} \right)+4{{\operatorname{Cos}}^{2}}\dfrac{x+y}{2}.{{\operatorname{Sin}}^{2}}\dfrac{x-y}{2} $ 

               $ =4{{\operatorname{Cos}}^{2}}\left( \dfrac{x+y}{2} \right)\left[ {{\operatorname{Cos}}^{2}}\dfrac{x-y}{2}+{{\operatorname{Sin}}^{2}}\dfrac{x-y}{2} \right] $ 

               $ =4{{\operatorname{Cos}}^{2}}\left( \dfrac{x+y}{2} \right) $ 

11. If  $ Cotx=-\dfrac{5}{12},x $  lies in second quadrant find the values of other five trigonometric functions 

 $ Cotx=-\dfrac{5}{12} $ 

Using some trigonometric identities, we obtain

 $ \tan x=-\dfrac{12}{5} $ 

 $ {{\operatorname{Sec}}^{2}}x=1+{{\tan }^{2}}x $ 

 $ \operatorname{Sec}x=\pm \dfrac{13}{5} $ 

Since  $ x $  lies in the second quadrant, the cosine value will be negative

 $ \operatorname{Sec}x=-\dfrac{13}{5} $ 

 $ \operatorname{Cos}x=-\dfrac{5}{13} $ 

 $ \operatorname{Sin}x=\tan x.\operatorname{Cos}x $ 

             $ =\dfrac{-12}{5}\times \left( \dfrac{-5}{13} \right) $ 

              $ =\dfrac{12}{13} $ 

 $ \operatorname{Csc}x=\dfrac{13}{12} $ 

12. Prove that  $ \dfrac{\operatorname{Sin}5x-2\operatorname{Sin}3x+\operatorname{Sin}x}{\operatorname{Cos}5x-\operatorname{Cos}x}=\tan x $  

Starting with the left-hand side and using the trigonometric difference identities for the sine function, we obtain

 $ \text{L}\text{.H}\text{.S}\text{.}=\dfrac{\operatorname{Sin}5x+\operatorname{Sin}x-2\operatorname{Sin}3x}{\operatorname{Cos}5x-\operatorname{Cos}x} $ 

                 $ =\dfrac{2\operatorname{Sin}3x.\operatorname{Cos}2x-2\operatorname{Sin}3x}{-2\operatorname{Sin}3x.\operatorname{Sin}2x} $ 

                 $ =\dfrac{2\operatorname{Sin}3x(\operatorname{Cos}2x-1)}{-2\operatorname{Sin}3x.\operatorname{Sin}2x} $ 

                 $ =\dfrac{-(1-\operatorname{Cos}2x)}{-\operatorname{Sin}2x} $ 

                 $ =\dfrac{2{{\operatorname{Sin}}^{2}}x}{2\operatorname{Sin}x.\operatorname{Cos}x} $ 

                 $ =\dfrac{\operatorname{Sin}x}{\operatorname{Cos}x} $ 

                 $ =\tan x $ 

13. Prove that  $ Sinx+Sin3x+Sin5x+Sin7x=4Cosx.Cos2x.Sin4x $  

Starting with the left-hand side and using the trigonometric addition identities for the sine function, we obtain

 $ \text{L}\text{.H}\text{.S}\text{.}=Sinx+Sin3x+Sin5x+Sin7x $ 

                 $ =\operatorname{Sin}x+\operatorname{Sin}7x+\operatorname{Sin}3x+\operatorname{Sin}5x $ 

                 $ =2\operatorname{Sin}\left( \dfrac{x+7x}{2} \right).\operatorname{Cos}\left( \dfrac{x-7x}{2} \right)+2\operatorname{Sin}\left( \dfrac{3x+5x}{2} \right)\operatorname{Cos}\left( \dfrac{3x-5x}{2} \right) $ 

                 $ =2\operatorname{Sin}4x.\operatorname{Cos}3x+2\operatorname{Sin}4x.\operatorname{Cos}x $ 

                 $ =2\operatorname{Sin}4x[\operatorname{Cos}3x+\operatorname{Cos}x] $ 

                 $ =2\operatorname{Sin}4x\left[ 2\operatorname{Cos}\left( \dfrac{3x+x}{2} \right).\operatorname{Cos}\left( \dfrac{3x-x}{2} \right) \right] $ 

                 $ =2\operatorname{Sin}4x[2\operatorname{Cos}2x.\operatorname{Cos}x] $ 

                 $ =4\operatorname{Cos}x.\operatorname{Cos}2x.\operatorname{Sin}4x $ 

14. Find the angle between the minute hand and hour hand of a clock when the time is  $ 7.20 $  

We know that the angle made by minute hand in  $ 15\min =15\times 6={{90}^{\circ }} $ 

We also know that the angle made by the hour hand in  $ 1hr={{30}^{\circ }} $ 

In  $ 60 $  minute  $ =\dfrac{30}{60} $ 

 $ =\dfrac{1}{2} $ 

 $ [\because  $ Angle Travelled by  $ hr $ hand in  $ 12hr={{360}^{\circ }}] $ 

In  $ 20 $ minutes  $ =\dfrac{1}{2}\times 20 $ 

                         $ ={{10}^{\circ }} $ 

Angle made  $ =90+10 $ 

                       $ ={{100}^{\circ }} $ 

15. Show that  $ \sqrt{2+\sqrt{2+2\operatorname{Cos}4\theta }}=2\operatorname{Cos}\theta  $  

Starting with the left-hand side and using the trigonometric addition identity for the cosine function, we obtain

 $ \text{L}\text{.H}\text{.S}\text{.}=\sqrt{2+\sqrt{2+2\operatorname{Cos}4\theta }} $ 

                 $ =\sqrt{2+\sqrt{2(1+\operatorname{Cos}4\theta )}} $ 

                 $ =\sqrt{2+\sqrt{2.2{{\operatorname{Cos}}^{2}}2\theta }} $ 

                 $ =\sqrt{2+2\operatorname{Cos}2\theta } $ 

                 $ =\sqrt{2(1+\operatorname{Cos}2\theta )} $ 

                 $ =\sqrt{2.2{{\operatorname{Cos}}^{2}}\theta } $ 

                 $ =2\operatorname{Cos}\theta  $ 

16. Prove that  $ Cot4x\left( Sin5x+Sin3x \right)=Cotx\left( Sin5xSin3x \right) $  

 $ \text{L}\text{.H}\text{.S}\text{.}=Cot4x\left( Sin5x+Sin3x \right) $ 

                $ =\dfrac{\operatorname{Cos}4x}{\operatorname{Sin}4x}\left[ 2\operatorname{Sin}\dfrac{5x+3x}{2}.\operatorname{Cos}\dfrac{5x-3x}{2} \right] $ 

                $ =\dfrac{\operatorname{Cos}4x}{\operatorname{Sin}4x}2\operatorname{Sin}4x.\operatorname{Cos}x $ 

                $ =2\operatorname{Cos}4x.\operatorname{Cos}x $ 

Then, we move on to the right-hand side and using the trigonometric addition identity for the sine function, we obtain

   $ \text{R}\text{.H}\text{.S}\text{.}=Cotx\left( Sin5xSin3x \right) $ 

                   $ =\dfrac{\operatorname{Cos}x}{\operatorname{Sin}x}\left[ 2\operatorname{Cos}\dfrac{5x+3x}{2}.\operatorname{Sin}\dfrac{5x-3x}{2} \right] $ 

                   $ =\dfrac{\operatorname{Cos}x}{\operatorname{Sin}x}[2\operatorname{Cos}4x.\operatorname{Sin}x] $ 

                   $ =2\operatorname{Cos}4x.\operatorname{Cos}x $ 

   $ \text{L}\text{.H}\text{.S}=\text{R}\text{.H}\text{.S} $ 

6 Marks Questions

1. Find the general solution of  $ sin2x+sin4x+sin6x=0 $ 

We have that  $ \sin 2x+\sin 4x+\sin 6x=0 $ 

 $ \Rightarrow \left( \sin 2x+\sin 6x \right)+\sin 4x=0 $ 

 $ \Rightarrow \left( 2\sin \left( \dfrac{2x+6x}{2} \right)\cos \left( \dfrac{2x-6x}{2} \right) \right)+\sin 4x=0 $ 

 $ \Rightarrow 2\sin 4x\cos 2x+\sin 4x=0 $ 

 $ \Rightarrow \sin 4x\left( 2\cos 2x+1 \right)=0 $ 

 $ \Rightarrow x=n\pi  $ 

 $ 2\cos 2x+1=0 $ 

 $ \Rightarrow x=n\pi \pm \dfrac{\pi }{3} $ 

2. Find the general solution of  $ \cos \theta \cos 2\theta \cos 3\theta =\dfrac{1}{4} $  

We have that  $ \cos \theta \cos 2\theta \cos 3\theta =\dfrac{1}{4} $ 

 $ \Rightarrow 4\cos \theta \cos 2\theta \cos 3\theta =1 $ 

Using the trigonometric addition identity for the cosine function, we obtain

 $ \Rightarrow 2\left( 2\cos \theta \cos 3\theta  \right)\cos 2\theta -1=0 $ 

 $ \Rightarrow 2\left( \cos 4\theta +\cos 2\theta  \right)\cos 2\theta -1=0 $ 

 $ \Rightarrow 2\left( 2{{\cos }^{2}}2\theta -1+\cos 2\theta  \right)\cos 2\theta -1=0 $ 

 $ \Rightarrow \left( 2{{\cos }^{2}}2\theta -1 \right)\left( 2\cos 2\theta +1 \right)=0 $ 

 $ 2{{\cos }^{2}}2\theta -1=0 $ 

 $ \Rightarrow \cos 4\theta =0 $ 

 $ \Rightarrow 4\theta =\left( 2n+1 \right)\dfrac{\pi }{2} $ 

 $ \Rightarrow \theta =\left( 2n+1 \right)\dfrac{\pi }{8} $ 

 $ 2\cos 2\theta +1=0 $ 

 $ \Rightarrow \cos 2\theta =-\dfrac{1}{2} $ 

 $ \Rightarrow \theta =n\pi \pm \dfrac{\pi }{3} $ 

3. If  $ \operatorname{Sin}\alpha +\operatorname{Sin}\beta =a $  and  $ \operatorname{Cos}\alpha +\operatorname{Cos}\beta =b $  

Show that  $ \operatorname{Cos}(\alpha +\beta )=\dfrac{{{b}^{2}}-{{a}^{2}}}{{{b}^{2}}+{{a}^{2}}} $  

Squaring both the equations and adding them together,

 $ {{b}^{2}}+{{a}^{2}}={{(\operatorname{Cos}\alpha +\operatorname{Cos}\beta )}^{2}}+{{(\operatorname{Sin}\alpha +\operatorname{Sin}\beta )}^{2}} $ 

                 $ ={{\operatorname{Cos}}^{2}}\alpha +{{\operatorname{Cos}}^{2}}\beta +2\operatorname{Cos}\alpha .\operatorname{Cos}\beta +{{\operatorname{Sin}}^{2}}\alpha +{{\operatorname{Sin}}^{2}}\beta +2\operatorname{Sin}\alpha .\operatorname{Sin}\beta  $ 

                 $ =1+1+2(\operatorname{Cos}\alpha .\operatorname{Cos}\beta +\operatorname{Sin}\alpha .\operatorname{Sin}\beta ) $ 

            $ =2+2\operatorname{Cos}(\alpha -\beta ) $        $ (1) $ 

 $ {{b}^{2}}-{{a}^{2}}={{(\operatorname{Cos}\alpha +\operatorname{Cos}\beta )}^{2}}-{{(\operatorname{Sin}\alpha +\operatorname{Sin}\beta )}^{2}} $ 

                 $ =({{\operatorname{Cos}}^{2}}\alpha -{{\operatorname{Sin}}^{2}}\beta )+({{\operatorname{Cos}}^{2}}\beta -{{\operatorname{Sin}}^{2}}\alpha )+2\operatorname{Cos}(\alpha +\beta ) $ 

                 $ =\operatorname{Cos}(\alpha +\beta )\operatorname{Cos}(\alpha -\beta )+\operatorname{Cos}(\beta +\alpha )\operatorname{Cos}(\alpha -\beta )+2\operatorname{Cos}(\alpha +\beta ) $ 

                 $ =2\operatorname{Cos}(\alpha +\beta ).\operatorname{Cos}(\alpha -\beta )+2\operatorname{Cos}(\alpha +\beta ) $ 

                 $ =\operatorname{Cos}(\alpha +\beta )[2\operatorname{Cos}(\alpha -\beta )+2] $ 

              $ =\operatorname{Cos}(\alpha +\beta ).({{b}^{2}}+{{a}^{2}}) $       from  $ (1) $ 

Dividing equation  $ \left( 1 \right) $  with  $ {{b}^{2}}+{{a}^{2}} $ , we get

 $ \dfrac{{{b}^{2}}-{{a}^{2}}}{{{b}^{2}}+{{a}^{2}}}=\operatorname{Cos}(\alpha +\beta ) $ 

4. Prove  $ Cos\alpha +Cos\beta +Cos\gamma +Cos\left( \alpha +\beta +\gamma  \right)=4\operatorname{Cos}\left( \dfrac{\alpha +\beta }{2} \right).\operatorname{Cos}\left( \dfrac{\beta +\gamma }{2} \right).\operatorname{Cos}\left( \dfrac{\gamma +\alpha }{2} \right) $ 

Starting with the left-hand side and using the trigonometric addition identities for the cosine function, we obtain

 $ \text{L}\text{.H}\text{.S}\text{.}=Cos\alpha +Cos\beta +Cos\gamma +Cos\left( \alpha +\beta +\gamma  \right) $ 

                 $ =2\operatorname{Cos}\left( \dfrac{\alpha +\beta }{2} \right).\operatorname{Cos}\left( \dfrac{\alpha -\beta }{2} \right)+2\operatorname{Cos}\left( \dfrac{\alpha +\beta +\gamma +\gamma }{2} \right).\operatorname{Cos}\left( \dfrac{\alpha +\beta +\gamma -\gamma }{2} \right) $ 

                $ =2\operatorname{Cos}\left( \dfrac{\alpha +\beta }{2} \right).\operatorname{Cos}\left( \dfrac{\alpha -\beta }{2} \right)+2\operatorname{Cos}\left( \dfrac{\alpha +\beta }{2} \right).\operatorname{Cos}\left( \dfrac{\alpha +\beta +2\gamma }{2} \right) $ 

                $ =2\operatorname{Cos}\left( \dfrac{\alpha +\beta }{2} \right)\left[ \operatorname{Cos}\left( \dfrac{\alpha -\beta }{2} \right)+\operatorname{Cos}\left( \dfrac{\alpha +\beta +2\gamma }{2} \right) \right] $ 

            $ =2\operatorname{Cos}\left( \dfrac{\alpha +\beta }{2} \right)\left[ 2\operatorname{Cos}\left( \dfrac{\dfrac{\alpha -\beta }{2}+\dfrac{\alpha +\beta +2\gamma }{2}}{2} \right).\operatorname{Cos}\left( \dfrac{\dfrac{\alpha +\beta +2\gamma }{2}-\dfrac{\alpha -\beta }{2}}{2} \right) \right] $ 

                $ =2\operatorname{Cos}\left( \dfrac{\alpha +\beta }{2} \right)\left[ 2\operatorname{Cos}\left( \dfrac{\alpha +\gamma }{2} \right).\operatorname{Cos}\left( \dfrac{\beta +\gamma }{2} \right) \right] $ 

               $ =4\operatorname{Cos}\left( \dfrac{\alpha +\beta }{2} \right).\operatorname{Cos}\left( \dfrac{\beta +\gamma }{2} \right).\operatorname{Cos}\left( \dfrac{\gamma +\alpha }{2} \right) $ 

5. Prove that  $ \operatorname{Sin}3x+\operatorname{Sin}2x-\operatorname{Sin}2x=4\operatorname{Sin}x.\operatorname{Cos}\dfrac{x}{2}.\operatorname{Cos}\dfrac{3x}{2} $  

 $ \text{L}\text{.H}\text{.S}\text{.}=\sin 3x+\sin x-\sin 2x $ 

                 $ =2\cos \left( \dfrac{3x+x}{2} \right).\operatorname{Sin}\left( \dfrac{3x+x}{2} \right)+\operatorname{Sin}2x $ 

                 $ =2\cos 2x.\sin x+\sin 2x $ 

                 $ =2\cos 2x.\sin x+2\sin x\cos x $ 

                 $ =2\sin x[\cos 2x+\cos x] $ 

                 $ =2\sin x\left[ 2\cos x\dfrac{3x}{2}.\cos \dfrac{x}{2} \right] $ 

                 $ =4\sin x\cos x\dfrac{3x}{2}\cos \dfrac{x}{2} $ 

6. Prove that  $ 2\cos \dfrac{\pi }{13}.\cos \dfrac{9\pi }{13}+\cos \dfrac{3\pi }{13}+\cos \dfrac{5\pi }{13}=0 $  

Starting with the left-hand side using the trigonometric addition identities for the cosine and sine function, we obtain

 $ \text{L}\text{.H}\text{.S}\text{.}=2\cos \dfrac{\pi }{13}.\cos \dfrac{9\pi }{13}+\cos \dfrac{3\pi }{13}+\cos \dfrac{5\pi }{13} $ 

                 $ =\cos \left( \dfrac{\pi }{13}+\dfrac{9\pi }{13} \right)+\cos \left( \dfrac{\pi }{13}-\dfrac{9\pi }{13} \right)+\cos \dfrac{3\pi }{13}+\cos \dfrac{5\pi }{13} $ 

                 $ =\cos \dfrac{10\pi }{13}+\cos \dfrac{18\pi }{13}+\cos \dfrac{3\pi }{13}+\cos \dfrac{5\pi }{13} $ 

                 $ =\cos \left( \pi -\dfrac{3\pi }{13} \right)+\cos \left( \pi -\dfrac{5\pi }{13} \right)+\cos \dfrac{3\pi }{13}+\cos \dfrac{5\pi }{13} $ 

                 $ =-\cos \dfrac{3\pi }{13}-\cos \dfrac{5\pi }{13}+\dfrac{3\pi }{13}+\cos \dfrac{5\pi }{13} $ 

7.  Find the value of  $ \tan (\alpha +\beta ) $ given that  $ \cot \alpha =\dfrac{1}{2},\alpha \in \left( \pi ,\dfrac{3\pi }{2} \right) $  and  $ \operatorname{Sec}\beta =-\dfrac{5}{3},\beta \in \left( \dfrac{\pi }{2},\pi  \right) $  

We know that,

 $ \tan \left( \alpha +\beta  \right)=\dfrac{\tan \alpha +\tan \beta }{1-\tan \alpha \tan \beta } $ 

 $ Cot\alpha =\dfrac{1}{2} $ 

 $ \tan \alpha =2 $ 

Now, let us find  $ \tan \beta  $ 

 $ 1+{{\tan }^{2}}\beta ={{\operatorname{Sec}}^{2}}\beta  $ 

 $ 1+{{\tan }^{2}}\beta ={{\left( \dfrac{-5}{3} \right)}^{2}}\left[ \because \operatorname{Sec}\beta =\dfrac{-5}{3} \right] $ 

 $ \tan \beta =\pm \dfrac{4}{3} $ 

 $ \tan \beta =-\dfrac{4}{3}\left[ \because \beta \in \left( \dfrac{\dfrac{\pi }{2}}{x} \right) \right] $ 

Therefore, we have that

 $ \tan \left( \alpha +\beta  \right)=\dfrac{2-\dfrac{4}{3}}{1-2\left( \dfrac{-4}{3} \right)} $ 

                     $ =\dfrac{2}{11} $ 

Prove that  $ \dfrac{\operatorname{Sec}8A-1}{\operatorname{Sec}4A-1}=\dfrac{\tan 8A}{\tan 2A} $  

Starting with the left-hand side and using the trigonometric elementary identities of the cosine function and sine function, we obtain

 $ \text{L}\text{.H}\text{.S}\text{.}=\dfrac{\sec 8A-1}{\sec 4A-1} $ 

                $ =\dfrac{\dfrac{1}{\operatorname{Cos}8A}-1}{\dfrac{1}{\operatorname{Cos}4A}-1} $ 

                $ =\dfrac{1-\operatorname{Cos}8A}{1-\operatorname{Cos}4A}\times \dfrac{\operatorname{Cos}4A}{\operatorname{Cos}8A} $ 

                $ =\dfrac{2{{\operatorname{Sin}}^{2}}4A}{2{{\operatorname{Sin}}^{2}}2A}.\dfrac{\operatorname{Cos}4A}{\operatorname{Cos}8A} $ 

                $ =\dfrac{(2\operatorname{Sin}4A.\operatorname{Cos}4A).\operatorname{Sin}4A}{2{{\operatorname{Sin}}^{2}}2A.\operatorname{Cos}8A} $ 

                $ =\dfrac{\operatorname{Sin}8A(2\operatorname{Sin}2A.\operatorname{Cos}2A)}{2{{\operatorname{Sin}}^{2}}2A.\operatorname{Cos}8A} $ 

                $ =\dfrac{\operatorname{Sin}8A\operatorname{Cos}2A}{\operatorname{Sin}2A.\operatorname{Cos}2A} $ 

                $ =\dfrac{\tan 8A}{\tan 2A} $ 

Prove that  $ {{\operatorname{Cos}}^{2}}x+{{\operatorname{Cos}}^{2}}\left( x+\dfrac{\pi }{3} \right)+{{\operatorname{Cos}}^{2}}\left( x-\dfrac{\pi }{3} \right)=\dfrac{3}{2} $  

Starting with the left-hand side and using trigonometric addition identities of the cosine function, we obtain

 $ \text{L}\text{.H}\text{.S}\text{.}=\dfrac{1+\operatorname{Cos}2x}{2}+\dfrac{1+\operatorname{Cos}\left( 2x+\dfrac{2\pi }{3} \right)}{2}+\dfrac{1+\operatorname{Cos}\left( 2x-\dfrac{2\pi }{3} \right)}{2} $ 

                $ =\dfrac{1}{2}\left[ 1+1+1+\operatorname{Cos}2x+\operatorname{Cos}\left( 2x+\dfrac{2\pi }{3} \right)+\operatorname{Cos}\left( 2x-\dfrac{2\pi }{3} \right) \right] $ 

                $ =\dfrac{1}{2}\left[ 3+\operatorname{Cos}2x+\operatorname{Cos}\left( 2x+\dfrac{2\pi }{3} \right)+\operatorname{Cos}\left( 2x-\dfrac{2\pi }{3} \right) \right] $ 

                $ =\dfrac{1}{2}\left[ 3+\operatorname{Cos}2x+2\operatorname{Cos}\left( \dfrac{2x+\dfrac{2\pi }{3}+2x-\dfrac{2\pi }{3}}{2} \right).\operatorname{Cos}\left( \dfrac{2x+\dfrac{2\pi }{3}-2x+\dfrac{2\pi }{3}}{2} \right) \right] $ 

                $ =\dfrac{1}{2}\left[ 3+\operatorname{Cos}2x+2\operatorname{Cos}2x.\operatorname{Cos}\dfrac{4\pi }{6} \right] $ 

                $ =\dfrac{1}{2}\left[ 3+\operatorname{Cos}2x+2\operatorname{Cos}2x.\operatorname{Cos}\dfrac{2\pi }{3} \right] $ 

            $ =\dfrac{1}{2}\left[ 3+\operatorname{Cos}2x+2\operatorname{Cos}2x.\operatorname{Cos}\left( \pi -\dfrac{\pi }{3} \right) \right] $ 

                $ =\dfrac{1}{2}\left[ 3+\operatorname{Cos}2x+2\operatorname{Cos}2x.\left( \dfrac{-1}{2} \right) \right] $ 

                $ =\dfrac{3}{2} $ 

 Prove that  $ \operatorname{Cos}2x.\operatorname{Cos}\dfrac{x}{2}-\operatorname{Cos}3x.\operatorname{Cos}\dfrac{9x}{2}=\operatorname{Sin}5x\operatorname{Sin}\dfrac{5x}{2} $  

Starting with the left-hand side and using trigonometric addition identities for the cosine function, we obtain

 $ \text{L}\text{.H}\text{.S}\text{.}=\dfrac{1}{2}\left[ 2\operatorname{Cos}2x.\operatorname{Cos}\dfrac{x}{2}-2\operatorname{Cos}3x.\operatorname{Cos}\dfrac{9x}{2} \right] $ 

                $ =\dfrac{1}{2}\left[ \operatorname{Cos}\left( 2x+\dfrac{x}{2} \right)+\operatorname{Cos}\left( 2x-\dfrac{x}{2} \right)-\operatorname{Cos}\left( \dfrac{9x}{2}+3x \right)-\operatorname{Cos}\left( \dfrac{9x}{2}-3x \right) \right] $ 

                $ =\dfrac{1}{2}\left[ \operatorname{Cos}\dfrac{5x}{2}+\operatorname{Cos}\dfrac{3x}{2}-\operatorname{Cos}\dfrac{15x}{2}-\operatorname{Cos}\dfrac{3x}{2} \right] $ 

               $ =\dfrac{1}{2}\left[ \operatorname{Cos}\dfrac{5x}{2}-\operatorname{Cos}\dfrac{15x}{2} \right] $ 

              $ =\dfrac{1}{2}\left[ -2\operatorname{Sin}\left( \dfrac{\dfrac{5x}{2}+\dfrac{15x}{2}}{2} \right).\operatorname{Sin}\left( \dfrac{\dfrac{5x}{2}-\dfrac{15x}{2}}{2} \right) \right] $ 

              $ =-\operatorname{Sin}5x.\operatorname{Sin}\left( \dfrac{-5x}{2} \right) $ 

              $ =\operatorname{Sin}5x.\operatorname{Sin}\dfrac{5x}{2} $ 

              $ =\text{R}\text{.H}\text{.S}\text{.} $ 

 Prove that  $ \operatorname{Cos}20{}^\circ .\operatorname{Cos}40{}^\circ .\operatorname{Cos}60{}^\circ .\operatorname{Cos}80{}^\circ =\dfrac{1}{16} $  

Starting with the left-hand side and using the trigonometric addition identities of the cosine function, we obtain

 $ \text{L}\text{.H}\text{.S}\text{.}=\operatorname{Cos}{{20}^{{}^\circ }}.\operatorname{Cos}{{40}^{{}^\circ }}.\operatorname{Cos}{{60}^{{}^\circ }}.\operatorname{Cos}{{80}^{{}^\circ }} $ 

                 $ =\operatorname{Cos}{{60}^{{}^\circ }}.\operatorname{Cos}{{20}^{{}^\circ }}.\operatorname{Cos}{{40}^{{}^\circ }}.\operatorname{Cos}{{80}^{{}^\circ }} $ 

                 $ =\dfrac{1}{2}.\dfrac{1}{2}\operatorname{Cos}{{40}^{{}^\circ }}\left( 2\operatorname{Cos}{{20}^{{}^\circ }}.\operatorname{Cos}{{80}^{{}^\circ }} \right) $ 

                 $ =\dfrac{1}{4}\operatorname{Cos}{{40}^{{}^\circ }}\left[ \operatorname{Cos}(80+20)+\operatorname{Cos}(80-20) \right] $ 

                 $ =\dfrac{1}{4}\operatorname{Cos}{{40}^{{}^\circ }}\left[ \operatorname{Cos}{{100}^{{}^\circ }}+\operatorname{Cos}{{60}^{{}^\circ }} \right] $ 

                $ =\dfrac{1}{4}\operatorname{Cos}{{40}^{{}^\circ }}\left[ \operatorname{Cos}{{100}^{{}^\circ }}+\dfrac{1}{2} \right] $ 

                $ =\dfrac{1}{8}(2\operatorname{Cos}{{100}^{{}^\circ }}\operatorname{Cos}{{40}^{{}^\circ }})+\dfrac{1}{8}\operatorname{Cos}{{40}^{{}^\circ }} $ 

           $ =\dfrac{1}{8}\left[ \operatorname{Cos}{{(100+40)}^{{}^\circ }}+\operatorname{Cos}{{(100-40)}^{{}^\circ }} \right]+\dfrac{1}{8}\operatorname{Cos}{{40}^{{}^\circ }} $ 

               $ =\dfrac{1}{8}\left[ \operatorname{Cos}{{140}^{{}^\circ }}+\operatorname{Cos}{{60}^{{}^\circ }} \right]+\dfrac{1}{8}\operatorname{Cos}{{40}^{{}^\circ }} $ 

               $ =\dfrac{1}{8}\left[ \operatorname{Cos}{{140}^{{}^\circ }}+\dfrac{1}{2} \right]+\dfrac{1}{8}\operatorname{Cos}{{40}^{{}^\circ }} $ 

              $ =\dfrac{1}{8}\operatorname{Cos}{{(180-40)}^{{}^\circ }}+\dfrac{1}{16}+\dfrac{1}{8}\operatorname{Cos}{{40}^{{}^\circ }} $ 

              $ =-\dfrac{1}{8}\operatorname{Cos}{{40}^{{}^\circ }}+\dfrac{1}{16}+\dfrac{1}{8}\operatorname{Cos}{{40}^{{}^\circ }} $ 

              $ =\dfrac{1}{16} $ 

             $ =\text{R}\text{.H}\text{.S}\text{.} $ 

 If  $ \tan x=\dfrac{3}{4},\pi <x<\dfrac{3\pi }{2}, $ Find the value of  $ \operatorname{Sin}\dfrac{x}{2},\operatorname{Cos}\dfrac{x}{2} $  and  $ \tan \dfrac{x}{2} $  

 $ \pi <x<\dfrac{3\pi }{2} $  implying that  $ x $  is in the third quadrant 

 $ \Rightarrow \dfrac{\pi }{2}<\dfrac{x}{2}<\dfrac{3\pi }{2} $ 

Therefore, we have that  $ \operatorname{Sin}\dfrac{x}{2} $  is positive and  $ \operatorname{Cos}\dfrac{x}{2} $  is negative.

Let us find for  $ \tan \dfrac{x}{2} $ 

 $ 1+{{\tan }^{2}}x={{\operatorname{Sec}}^{2}}x\dfrac{5}{4} $ 

 $ 1+{{\left( \dfrac{3}{4} \right)}^{2}}={{\operatorname{Sec}}^{2}}x $ 

    $ {{\operatorname{Sec}}^{2}}x=\pm \dfrac{25}{16} $ 

      $ \operatorname{Cos}x=\pm \dfrac{4}{5} $ 

      $ \operatorname{Cos}x=-\dfrac{4}{5}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ \because \pi <x<\dfrac{3\pi }{2} \right] $ 

Let us find the required values 

 $ \operatorname{Sin}\dfrac{x}{2}=\sqrt{\dfrac{1-\operatorname{Cos}x}{2}} $ 

              $ =\sqrt{\dfrac{1+\dfrac{4}{5}}{2}} $ 

              $ =\sqrt{\dfrac{9}{10}} $ 

              $ =\dfrac{3}{\sqrt{10}} $ 

  $ \operatorname{Cos}\dfrac{x}{2}=-\sqrt{\dfrac{1-\operatorname{Cos}x}{2}} $ 

                  $ =-\sqrt{\dfrac{1-\dfrac{4}{5}}{2}} $ 

                  $ =-\sqrt{\dfrac{1}{10}} $ 

                 $ =\dfrac{-1}{\sqrt{10}} $ 

  $ \tan \dfrac{x}{2}=\dfrac{\dfrac{3}{\sqrt{10}}}{\dfrac{-1}{\sqrt{10}}} $ 

                $ =-3 $ 

Class 11 Maths Chapter 3 Important Questions- What are Trigonometric Functions? 

In simple language, the trigonometric functions are the functions of an angle of triangles. It defines the relationship between sides and angles of a triangle is given on basis of these functions. The trigonometric functions consist of the sine, cosine, sectant, cosecant, tangent, and cotangent. It is also known as circular functions. There are several trigonometric formulas and identities that help to define the relationship between the angles and also the functions. At the end of chapter 3, students will find trigonometric functions class 11 extra questions. 

Tips to Score Marks in Trigonometric Functions?

The trigonometric function is one of the important chapters of class 11 maths. The concept of trigonometry was mainly developed to solve geometric problems that incorporate triangles. By practising the important questions of maths class 11 trigonometry functions, students can easily score high marks in the examinations. When students prepare these important questions from Vedantu, they can also learn several tricks and shortcuts to solve the questions fast. Besides, students need to focus on the formulas of trigonometric functions that are crucial to solve the sums. Students should not skip this chapter at any cost as there are many significant areas like designing electronic circuits, finding the heights of tides, etc. To get a deeper insight into class 11 maths ch 3 important questions, students should practice the resources available from Vedantu. 

Discuss the Trigonometric Tables and Formulas?

In class 11 trigonometric functions, important questions about trigonometric tables and formulas constitute a vital part of the chapter. Let’s discuss both these concepts in detail below. 

The Formula for Function of Trigonometric Ratios

Trigonometric table, important questions for class 11 maths chapter 3 based on exercise.

Q. An engine produces 360 revolutions in one minute. Through how many radians will it turn in one second?

The total number of revolutions made by an engine in one minute = 360

1 minute = 60 seconds

Therefore, number of revolutions in 1 second = 360/60 = 6

Angle formed in 1 revolution = 360°

Angles formed in 6 revolutions = 6 × 360°

Radian measure of the angle in a total of six revolutions = 6 × 360 × π/180

= 6 × 2 × π

So, the engine turns 12π radians in one second.

Importance of Downloading Class 11 Maths Chapter 3 Important Questions PDF?

By downloading the important questions for class 11 maths chapter 3 students will get exposure to the concept of trigonometric functions in depth. Here are some benefits that students will get when they have the PDF version.

They can prepare important notes for the examination. 

They will get access to trigonometry functions class 11 extra questions. 

They can use it as a ready resource for reference. 

It will help them understand the question pattern of the examination. 

Practice Questions

Prove that: (sin 5x + sin 3x)/(cos 5x + cos 3x) = tan 4x

Find the value of tan 765° cot 675° + tan 225° cot 405°

Solve the equation: tan² θ + cot² θ = 2

Write the value of 2sin 75° sin 15°

Show that: tan 4A = (cos8Acos5A - cos12Acos9A) / (sin8Acos5A + cos12Asin9A)

Find the general solution of the following equation: tan2θ +(1 – √3) tan θ – √3 = 0

Prove that: 3sinπ/6secπ/3 - 4sin5π/6cotπ/4 = 1

Find the value: cos 4 π /8 + cos 4 3 π /8 + cos 4 5 π /8 + cos 4 7π/8

Show that: tan 15° + cot 15° = 4

Find the most general value of θ satisfying the equation tan θ = -1 and cos θ = 1/√2

Advantages of Opting Vedantu for Important Questions of Class 11 Maths: Chapter 3 Trigonometric

Vedantu offers several benefits to students using their platform for the "Important Questions for CBSE Class 11 Maths Chapter 3 - Trigonometric Functions (2023-24)":

Comprehensive Coverage: Vedantu's important questions are curated to cover a wide spectrum of topics within the chapter, ensuring a comprehensive understanding of trigonometric functions.

Strategic Exam Preparation : These questions are strategically selected to align with the CBSE curriculum and examination patterns, preparing students effectively for their exams.

Conceptual Clarity: Vedantu's platform emphasizes conceptual clarity by providing in-depth explanations and solutions for each question, helping students grasp the underlying principles.

Variety of Problem Types : The diverse range of questions offered by Vedantu challenges students to apply trigonometric concepts in various problem-solving scenarios, enhancing their problem-solving skills.

Self-Assessment and Practice: Students can use these questions for self-assessment and regular practice, enabling them to gauge their progress and identify areas that need improvement.

Flexibility and Convenience: Vedantu's platform allows students to access these questions anytime, anywhere, providing flexibility in their study routine.

Personalized Learning: Vedantu's adaptive learning approach tailors the learning experience to each student's pace and needs, ensuring effective comprehension and retention.

Interactive Sessions: Vedantu offers interactive live sessions where students can clarify doubts and seek guidance from experienced educators, ensuring a holistic learning experience.

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Holistic Support: Beyond just important questions, Vedantu provides additional study material, revision notes, and comprehensive study plans to support students' overall exam preparation.

Prepare Well with Vedantu Important Questions for CBSE Class 11 Maths Chapter Trigonometric Function 2023-24

The compilation of important questions for CBSE Class 11 Maths Chapter 3 - "Trigonometric Functions" (2023-24) offers a strategic tool to enhance students' grasp of fundamental trigonometric concepts. These questions delve into various aspects of trigonometry, aiding in the development of problem-solving skills and conceptual clarity. By addressing a diverse range of problem types, these questions prepare students comprehensively for examinations and instil a deeper understanding of the subject matter. This resource not only facilitates exam preparation but also promotes a robust foundation for more advanced mathematical exploration, ensuring students' competence in applying trigonometric principles in various scenarios.

Important Related Links for CBSE Class 11 

FAQs on Important Questions for CBSE Class 11 Maths Chapter 3 Trigonometric Functions 2023-24

Q1. What is the easiest way to learn Trigonometry of Class 11 Maths?

Ans: The easiest way to learn Trigonometry is by practice and hard work. Students need to learn all the formulas to solve the questions that might be asked in the exams.  Therefore, the first thing they need to do in the chapter of Trigonometry is to memorize all the formulas by heart and revise them repeatedly, so that it stays in their memory. After this, students must practise all the important questions provided by Vedantu. This will give them an idea of how the question might be asked in the exams. The questions asked in the question paper are never straightforward. Thus, students must have a strong understanding of the concepts to do well in the Class 11 Maths exam.

Q2. Is Trigonometry important for JEE?

Ans: Yes. Trigonometry is one of the most important and prominent topics on which questions are set in the JEE main. Trigonometry functions and Trigonometry ratios are some of the important areas that you might need to focus on. Thus, a good hold on the Trigonometry of Class 11 not only helps students to score more than 90 in their Maths exam but also prepares them for other competitive exams that will shape their future career. Therefore, it goes without saying that students need to practice and work hard in the concepts of Chapter 3 of Class 11 Maths. The NCERT exercises must be practised regularly, to get a better hold of the concepts.

These solutions are available on Vedantu's official website( vedantu.com ) and mobile app free of cost.

Q3. How do you solve Chapter 3 Trigonometry of Class 11 Maths?

Ans: To solve all the numerical problems that might be asked based on the chapter on Trigonometry, students must memorize all the formulas. This will help them solve any question that might be asked in the question paper based on the chapter on Trigonometry. In addition to this, they should practice all the exercises from the NCERT Class 11 Maths Chapter 3. This will prepare and groom them to analyze and solve the questions.

Q4. Is Trigonometry of Class 11 Maths hard?

Ans: Chapter 3 Trigonometry of Class 11 Maths may seem challenging for some students as it has a lot of formulae and derivations to memorize. To score more than 90 in the Class 11 Maths exam, students must practice and revise the important concepts of Chapter 3 regularly. They should also figure out their areas of weaknesses from the chapter and work extra hard on these. Practice is the most important factor to crack this chapter.

Q5. Are the Important Questions for Chapter 3 of Class 11 Maths helpful?

Ans: Vedantu provides Important Questions for Chapter 3 of Class 11 Maths. These are extremely helpful for students as they can help form the base for Class 12. These questions are prepared by experts at Vedantu for the benefit of students. There is a high chance that these questions could be asked in the Class 11 Maths exam. Hence, students should make sure to solve them while they are preparing for their Maths exam, to get a hang of the concepts and the question pattern.

CBSE Class 11 Maths Important Questions

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CASE STUDY-1 CHAPTER -1 SET THEORY

a) 2 b) 6 c) 20 d) None of these

a) 9 b) 6 c) 1 d) 3

Q 6) French and English but not Sanskrit a) 2 b) 4 c) 6 d) 12

CASE STUDY-2 CHAPTER -1 SET THEORY

Case study-1 chapter -2 relations & functions.

a) {(1, 0), (1, 1), (1, 2), (2, 0), (2, 1), (2, 2), (3, 0), (3, 1), (3, 2), (4, 0), (4, 1), (4, 2)

b) {(1, 1), (1, 2), (2, 1), (2, 2), (3, 1), (3, 2),  (4, 1), (4, 2), (5, 1), (5, 2)}

c) {(1, 1), (1, 2), (2, 1), (2, 2), (3, 1), (3, 2), (4, 1), (4, 2)

d) {(1, 0), (1, 1), (1, 2), (2, 0), (2, 1), (2, 2), (3, 0), (3, 1), (3, 2), (4, 0), (4, 1), (4, 2), (5, 0), (5, 1),(5, 2)

b) {(0, 1), (0, 2),  (1, 1), (1, 2), (2, 1), (2, 2), (3, 1), (3, 2), (4, 1), (4, 2)

c) {(1, 0), (1, 1), (1, 2), (2, 0), (2, 1), (2, 2), (3, 0), (3, 1), (3, 2), (4, 0), (4, 1), (4, 2), (5, 0), (5, 1), (5, 2)

d) {(1, 1),  (1, 2), (2, 1), (2, 2), (3, 1), (3, 2), (4, 1), (4, 2)

R = {(x, y): x + y = 4, x   ∈  A, y  ∈  B} as a set of ordered pair is

a) R = {(1, 3), (3, 1), (0, 4), (4, 0)

b)R = {(2, 2), (3, 1), (1, 3)

c) R = {(2, 2), (3, 1), (4, 0)}

d) R = {(2, 2), (1, 3), (0, 4)}

a) {0, 1, 3, 4}                  

b) {1, 2, 3}         

c) 2, 3, 4}         

d) {0, 1, 2}

a) {0, 1, 2}           

b) {2, 3, 4}          

c) {1, 2, 3}             

d) {0, 1, 3, 4}

CASE STUDY-2 CHAPTER -2 RELATIONS & FUNCTIONS

CASE STUDY-1 CHAPTER -6 LINEAR INEQUALITIES

Marks obtained by Radhika in quarterly and half yearly examinations of Mathematics are 60 and 70 respectively.

Based on the above information, answer the following questions

a)  80               

b) 85                  

c) 75                  

d)  90

a)  85              

b) 90                  

c) 95                  

a)  [60, 70]            

b) [50, 80]                 

c) [50, 70]             

d) [60, 80]

 a) [50, 70]            

b) [60, 70]                

c) 50, 60]                 

CASE STUDY-1 CHAPTER -7 

Permutations & combinations.

Q3) How many different telephone numbers are there in each zone with all digits distinct?

CASE STUDY-2 CHAPTER -7 

Case study-1 chapter -9 , sequence & series.

Arithmetic Progression is a sequence in which the difference of any two consecutive terms remain same throughout the sequence. Above figures are made up of squares and the count of these squares are in AP. Carefully observe above figures.

Based on the above information answer the following questions

a)  17                  b)  31                  c)  34                  d) 37

a)  23rd                b)  24th                     c)  25th                     d)  26th

a)  9.75 cm 2                      b) 10 cm 2                          c)  10.75 cm 2                         d) 11 cm 2

a)  67                  b)  85                  c)  93                  d)  139

CASE STUDY-2 CHAPTER -9 

CASE STUDY-3 CHAPTER -9 

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Syllabus for the Session 2023-24

CBSE Syllabus

Case Study Questions

Case Study on Sets      CS-2   CS-3   CS-4   CS-5

Case Study on Relations & Functions   CS-2   CS-3

Case Study on Trigonometric Functions

Case Study on Complex Numbers

Case Study on Linear Inequalities

Case Study on Permutations and Combinations

Case Study on Sequences & Series

Case Study on Straight Lines

Case Study on Conic Sections

Case Study on Statistics

Case Study on Probability

Pdf of  Case Studies

MCQs for Practice

Chapter 1 - Sets     

Chapter 2 - Relations & Functions

Chapter 3 - Trigonometric Functions

Chapter 4 - Complex Numbers & Quadratic Equations

Chapter 5 - Linear Inequalities

Chapter 6 - Permutations & Combinations

Chapter 7 - Binomial Theorem

Chapter 8 - Sequences & Series

Chapter 9 - Straight Line

Chapter 1 0 - Conic Sections

Chapter 1 1 - Introduction to Three-dimensional Geometry

Chapter 12 - Limits & D erivatives

Chapter 13 - Statistics

Chapter 14 - Probability

Answers of MCQs

Assertion & Reasoning Questions

Relations & Functions

Trigonometric Functions

Complex Numbers

Linear Inequalities

Permutations & Combinations

Binomial Theorem

Sequences & Series

Straight Lines

Conic Sections

Introduction to Three-Dimensional Geometry

Limits & derivatives

Probability

Topic Wise Assignments of Previous Year Questions

Relations and Functions

Straight Line

Limits & Derivatives

Question Papers - DoE, Delhi

Session 2015-2016

            SA2(QP)         SA2(MS)         COMP(QP)         COMP(MS)

Session 201 6 -20 17

            SA1(QP)      SA2(QP)         SA2(MS)         COMP(QP)        COMP(MS)

Session 201 7 -201 8

            SA2(QP)          SA2(MS)         COMP(QP)         COMP(MS)

Session 201 8 -201 9

            SA2(QP)          SA2(MS)         COMP(QP)         COMP(MS)

Session 201 9 -20 20

            SA2(QP)          SA2(MS)        

Session 2020-21

    Video Explanation:  Part A     Part B Section III     Part B Section IV   Part B Section V

Session 2021-22

Question Paper 

Video Explanation:  Section A     Section B     Section C

Session 2022-23

  Sample Question Paper for Mid-Term Exam

Video Explanation: Section A     Section B    Section C     Section D     Section E

Mid-Term Exam:  Question Paper

Video Explanation: Section A     Section B     Section C     Section D    Section E

Practice Paper for Final Exam

Video Explanation: Section A     Section B     Section C     Section D     Section E

Final Exam:  Question Paper     Marking Scheme

Video Explanation: Section A     Section B     Section C     Section D     Section E

Support Material Issued by DoE, Delhi

Session 202 3 -24

 Mid-Term Exam:  Question Paper

Video Explanation:   Section A    Section B     Section C     Section D     Section E

Practice Paper 1 for Final Exam: Question Paper

Practice Paper 2 for Final Exam: Question Paper

Short Capsules (Notes to Revise Concepts)

Chapter 1 - Sets

Chapter 4 - Mathematical Induction

Chapter 5 - Complex Numbers & Quadratic Equations

Chapter 6 - Linear Inequalities

Chapter 7 - Permutations & Combinations

Chapter 8 - Binomial Theorem

Chapter 9 - Sequences & Series

Chapter 10 - Straight Line

Chapter 11 - Conic Sections

Chapter 12 - Introduction to Three-dimensional Geometry

Chapter 13 - Limits & derivatives

Chapter 14 - Mathematical Reasoning

Chapter 15 - Statistics

Chapter 16 - Probability

Some Basic Concepts to Revise:

Number System

Mensuration Results

Quadratic Equations

Solution of Polynomial Inequality - Wavy Curve Method

Probability__Revision of the Topics studied in Earlier Classes  

Result Sheets

Trigonometric Identities

Study Material

Home > Class 11 Maths Important Questions

Chapter 3 Trigonometric Functions Class 11 Maths Important Questions

Trigonometric Functions Class 11 Important Questions contains various topics that are asked every year in final examinations. In this chapter, you will understand various topics like measurement of angles, sexagesimal system, centisimal system, trigonometric ratios, radian system, etc. 

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CBSE Chapter 3 Trigonometric Functions HOTs Questions for Class 11 PDF

CBSE Chapter 3 Trigonometric Functions HOTs Questions for Class 11 download here at free of cost. you can check here CBSE Chapter 3 Trigonometric Functions HOTs Questions for Class 11 based on latest syllabus and examination pattern. CBSE Chapter 3 Trigonometric Functions HOTs Questions helps to improve conceptual understanding and develops thinking skills. CBSE Chapter 3 Trigonometric Functions HOTs Questions and answers for Class 11 are available for free download in pdf format.

CBSE Chapter 3 Trigonometric Functions HOTs Questions for Class 11

CBSE Board introduced questions based on - Higher Order Thinking Skills (HOTs). The primary objective here was to improve the evaluating and analytical skills of a student. A lot of students tend to only focus on memorizing information and relying on cramming learning. Therefore CBSE Board decided to dedicate a large part of the question paper to Chapter 3 Trigonometric Functions HOTs Questions as they typically concentrate on a student's ability to reason, analyze, process, justify,  and evaluate information. Since a large part of CBSE question papers is based on HOTS, it is necessary for students to get ample practice with these questions. Experts at SelfStudys have compiled HOTs question for Class 11 for Science and Maths. 

CBSE releases Chapter 3 Trigonometric Functions HOTs Questions for Class 11 every year. These papers are published prior to the examinations so that students can do the study. The students should practice Chapter 3 Trigonometric Functions HOTs Questions to increase perfection which will help him to get good marks in CBSE examination. Based on CBSE and NCERT guidelines same pattern as released every year.

In simple words, CBSE Class 11 HOTs Questions are the skills that involve learning that goes beyond mere comprehension. Higher Order Thinking Skills (HOTs) as the name suggests refers to thinking skills that correspond to the more complicated forms of learning within the cognitive domain. Mathematics HOTS Question HOTs Questions involve behaviour ranging from the simplest to the most difficult. 

CBSE Chapter 3 Trigonometric Functions HOTs Questions which are basically designed to improve the learning skill of students and test their comprehension, analytical thinking and problem-solving skill. In CBSE board exams, most of the questions which are of higher difficulty level are taken from NCERT books. So students must widely practice all the CBSE Chapter 3 Trigonometric Functions HOTs Questions so that they can easily attempt even the difficult questions asked in board examinations. While practising the CBSE Chapter 3 Trigonometric Functions HOTs Questions, students must keep a reliable source to seek precise solutions so that they may know the right way to approach the answers to Mathematics HOTS Question HOTs Questions. In this page, we are providing the CBSE Class 11 HOTs Questions for two major subjects such as Maths and Science. All the CBSE Chapter 3 Trigonometric Functions HOTs Questions provided here are prepared by experts to bring accurate study material for Class 11 students. All these CBSE Chapter 3 Trigonometric Functions HOTs Questions are designed in a step-wise manner to provide an easy and simple explanation so that students can easily learn the concepts and facts used.

Understanding of CBSE Chapter 3 Trigonometric Functions HOTs Questions for Class 11

Class 11 requires a strategic approach and the CBSE Chapter 3 Trigonometric Functions HOTs Questions provided on our website for every chapter form a cornerstone for thorough preparation. Class 11 is one of the most important stages in students life. This Chapter 3 Trigonometric Functions HOTs Questions can help to get a high overall score and hence needs dedicated preparation.

In the run-up to board exams of Class 11, it is important to build the base itself. Getting a good mark in Class 11 examinations will not only boost your confidence but will also give you the motivation to study for the next class. In addition, our website also has important questions for Class 11 subject wise solutions which will help students to refer to the answers once they are done answering in order to self-evaluate their performance and correct any errors.

Our aim is to make this CBSE Mathematics HOTS Question available to every student. This will enable you to practice the subject better and get high marks in the exam. Practising these Chapter 3 Trigonometric Functions HOTs Questions is a good way to ensure that you have not left any topic and that you have complete coverage of the whole syllabus. Also, students can use these Mathematics HOTS Question to take a mock test and use it as a tool to gauge their preparation. The Chapter 3 Trigonometric Functions HOTs Questions and answers are developed by referring to previous year question papers given in the NCERT textbooks. The format of the answers is in accordance with the latest recommended format given by the CBSE. This allows students to study the answer and replicate it in the examination in similar language and get good marks.

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Unit 3: Trigonometric functions

Degrees and radians.

• Intro to radians (Opens a modal)
• Radians & degrees (Opens a modal)
• Degrees to radians (Opens a modal)
• Radians to degrees (Opens a modal)
• Radian angles & quadrants (Opens a modal)
• Radians & degrees Get 3 of 4 questions to level up!

Unit circle

• Unit circle (Opens a modal)
• The trig functions & right triangle trig ratios (Opens a modal)
• Trig values of π/4 (Opens a modal)
• Trig unit circle review (Opens a modal)
• Plotting angles on unit circle Get 3 of 4 questions to level up!
• Sign of trigonometric functions Get 3 of 4 questions to level up!
• Trig values of special angles Get 3 of 4 questions to level up!

Trigonometric functions

• Graph of y=sin(x) (Opens a modal)
• Graph of y=tan(x) (Opens a modal)
• Intersection points of y=sin(x) and y=cos(x) (Opens a modal)
• Find value of other trigonometric functions from given trigonometric function Get 3 of 4 questions to level up!

Trigonometric identities: Symmetry

• Sine & cosine identities: symmetry (Opens a modal)
• Tangent identities: symmetry (Opens a modal)
• Sine & cosine identities: periodicity (Opens a modal)
• Tangent identities: periodicity (Opens a modal)

Trigonometric identities: Sum and difference

• Trig angle addition identities (Opens a modal)
• Using the cosine angle addition identity (Opens a modal)
• Using the cosine double-angle identity (Opens a modal)
• Proof of the sine angle addition identity (Opens a modal)
• Proof of the cosine angle addition identity (Opens a modal)
• Trig challenge problem: cosine of angle-sum (Opens a modal)
• Trigonometric functions of sum and difference of angles Get 3 of 4 questions to level up!
• Sum and difference of trigonometric functions Get 3 of 4 questions to level up!
• Evaluate trigonometric expressions (intermediate) Get 3 of 4 questions to level up!

Trigonometric equations

• Proof of the Pythagorean trig identity (Opens a modal)
• Using the Pythagorean trig identity (Opens a modal)
• Solving sinusoidal equations of the form sin(x)=d (Opens a modal)
• Solving cos(θ)=1 and cos(θ)=-1 (Opens a modal)
• Use the Pythagorean identity Get 3 of 4 questions to level up!
• Principal solutions of trigonometric equation Get 3 of 4 questions to level up!
• General solution of trigonometric equation Get 3 of 4 questions to level up!

Solutions to select NCERT problems

• Select problems from exercise 3.3 (Opens a modal)
• Select problems from miscellaneous exercise (Opens a modal)

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Right to Self-Determination Class 11 Polity Notes

Nations aspire for self-governance and shaping their future, seeking autonomy in decision-making and development pathways. Self-determination encompasses the pursuit of acknowledgment as a distinct political entity, reflecting the collective will of a community. Rooted in deep-seated communal identities and historical narratives, the concept transcends geographical and temporal boundaries. Both a historical and contemporary phenomenon, self-determination movements continue to shape global politics and international relations.

Right to Self-Determination

• Nations seek acknowledgment and validation of their sovereignty on the international stage, striving for legitimacy and recognition.
• The concept is intricately linked with cultural preservation and empowerment, representing a struggle for identity and autonomy.
• Historical precedents, particularly evident in post-World War I Europe, underscore the complexities of realizing self-determination.
• Evolving from the ideal of “one culture – one state,” the concept encompasses diverse interpretations and implementations across regions.
• Struggles to reconcile cultural autonomy with the establishment of viable state structures highlight the nuanced nature of self-determination.
• Challenges abound in accommodating diverse ethnicities within newly drawn borders, often resulting in tensions and conflicts.
• The quest for political recognition extends beyond sovereignty, encompassing the protection of minority rights and cultural heritage.
• At its core, the right to self-determination embodies aspirations for dignity, equality, and self-expression.

Challenges and Paradoxes

• The reorganization of boundaries in pursuit of self-determination often precipitates mass displacement and communal violence.
• Newly formed states grapple with the complexities of multiculturalism, confronting the realities of pluralistic societies within their borders.
• Minority communities within nation-states face systemic marginalization and discrimination, highlighting the tensions between unity and diversity.
• Balancing political autonomy with the protection of minority rights poses significant governance challenges for emerging nations.
• The unintended consequences of post-colonial state formation continue to reverberate, perpetuating historical injustices and conflicts.
• Historical context plays a pivotal role in shaping contemporary challenges, underscoring the enduring legacies of colonialism and imperialism.
• Inclusive governance structures that respect diversity and promote social cohesion are imperative for addressing the paradoxes of self-determination.
• Resolving the tensions between national sovereignty and minority rights remains a central dilemma in the pursuit of inclusive nation-building.

National Liberation Movements

• National liberation movements emerge in response to colonial domination, advocating for political independence and collective empowerment.
• Driven by aspirations for dignity, recognition, and self-governance, these movements represent a quest for justice and liberation.
• The legacy of migration, border disputes, and conflict underscores the complexities of decolonization and post-colonial state formation.
• Success rates in achieving political independence vary, with some movements achieving statehood while others face ongoing struggles for recognition.
• Cultural and linguistic preservation are central to the identity of oppressed communities, fueling resistance against colonial assimilation.
• Global decolonization movements catalyze solidarity among oppressed peoples, fostering international support for liberation struggles.
• The intersectionality of national liberation movements with human rights and global justice movements underscores their broader significance.
• The legacies of colonialism continue to shape present-day geopolitics, influencing patterns of conflict, cooperation, and identity politics.

Contemporary Dilemmas

• Self-determination movements persist as a pervasive force in global politics, challenging established state boundaries and governance structures.
• The complexities of statehood in a globalized world underscore the need for flexible and inclusive approaches to nation-building.
• A shift towards democratic inclusivity over separatism reflects evolving norms and values in contemporary statecraft.
• Multicultural coexistence within nation-states requires nuanced approaches to governance that respect diversity and promote social cohesion.
• Striking a balance between national unity and minority rights is essential for fostering inclusive and resilient societies.
• International norms and standards play a crucial role in shaping responses to self-determination movements, influencing diplomatic and legal frameworks.
• The implications of self-determination movements for global peace and security highlight the interconnectedness of local and global politics.
• Dialogue and diplomacy are essential tools for managing conflicts arising from self-determination movements, fostering peaceful resolutions and reconciliation.

Case Study: Basque Nationalism

• The Basque region’s quest for nationhood within Spain epitomizes the complexities of self-determination.
• Rooted in a distinct cultural and linguistic identity, Basque nationalism reflects centuries-old aspirations for autonomy and self-governance.
• Historical resistance against Spanish rule shapes modern Basque nationalist movements, fueling ongoing struggles for recognition and autonomy.
• Socio-political dynamics within Basque society inform divergent perspectives on the legitimacy of separatist aspirations.
• Contrasting narratives surrounding Basque identity and autonomy highlight the complexities of historical memory and political discourse.
• The implications of Basque nationalism extend beyond regional dynamics, impacting Spanish national cohesion and identity.
• The Basque case underscores the importance of understanding historical and cultural contexts in evaluating self-determination movements.
• Dialogue and negotiation remain essential tools for addressing the grievances and aspirations of Basque nationalists within the framework of Spanish national unity.
• Self-determination embodies the aspirations of communities for autonomy, dignity, and self-expression.
• Addressing the challenges and paradoxes of self-determination requires nuanced and inclusive approaches to governance.
• Balancing national sovereignty with minority rights is essential for fostering resilient and inclusive societies.
• Learning from historical struggles informs contemporary responses to self-determination movements, promoting dialogue and reconciliation.
• Upholding human rights and dignity amidst cultural diversity is essential for building peaceful and just societies.
• The quest for self-determination is an ongoing process, shaped by historical legacies, contemporary dynamics, and evolving norms and values.
• Dialogue, diplomacy, and mutual respect are essential for navigating the complexities of self-determination in a globalized world.
• Striving for inclusive nation-building entails recognizing and honoring the diversity of identities, aspirations, and experiences within and across borders.

Related Links

• CBSE Sample Papers for Class 11 Political Science (2023-24) Set 1 with Solutions
• CBSE Sample Papers for Class 11 Political Science Set 2 with Solutions
• Fundamental Rights And Writs Of India

Right to Self-Determination Class 11 Polity Notes- FAQs

What is the utility of the right of national self-determination.

“All peoples have the right freely to determine, without external interference, their political status and to pursue their economic, social and cultural development, and every State has the duty to respect this right in accordance with the provisions of the Charter”.

What is self-determination?

Self-determination is a combination of skills, knowledge, and beliefs that enable a person to engage in goal-directed, self-regulated, autonomous behavior. 

What are the main principles behind self-determination?

In a practical sense, self-determination means that Indigenous peoples have the freedom to live well and humanly, to determine what it means to live humanly, and to live according to our own values and beliefs.

What are the limitations of self-determination?

One limitation is that the theory does not account for the social and cultural factors that influence motivation and behavior. 

What are the essential elements of self-determination?

Components of Self-Determination Autonomy: People need to feel in control of their own behaviors and goals. … Competence: People need to gain mastery of tasks and learn different skills. … Connection or relatedness: People need to experience a sense of belonging and attachment to other people.

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Class 11 Maths Chapter 3 Trigonometric Functions MCQs

Class 11 Maths Chapter 3 Trigonometric Functions MCQs are available here for the students to help them score good marks in the board exam 2022-2023. Here, you will get the objective type questions on trigonometric functions along with correct options and explanations. These multiple-choice questions help you in practising a variety of questions on Chapter 3 of Class 11 maths.

Get MCQs for all the chapters of Class 11 Maths here.

MCQs for Chapter 3 Trigonometric Functions

Students are advised to practise the multiple-choice questions on Chapter 3 of Class 11 maths to understand how to apply the formulas of trigonometry.

Class 11 Maths Chapter 3 Trigonometric Functions – Download PDF

MCQs of Class 11 Maths Chapter 3 covers all the concepts of the NCERT curriculum. These MCQs will help you to improve your problem-solving skills and boost your confidence.

Also, check:

• Trigonometric Functions Class 11 Notes
• Important Questions for Class 11 Maths Chapter 3 Trigonometric Functions

MCQs for Chapter 3 Trigonometric Functions Class 11 with Answers

1. If sin θ and cos θ are the roots of ax 2 – bx + c = 0, then the relation between a, b and c will be

(a) a 2 + b 2 + 2ac = 0

(b) a 2 – b 2 + 2ac = 0

(c) a 2 + c 2 + 2ab = 0

(d) a 2 – b 2 – 2ac = 0

Correct option: (b) a 2 – b 2 + 2ac = 0

Given that sin θ and cos θ are the roots of the equation ax 2 – bx + c = 0, so sin θ + cos θ = b/a

and sin θ cos θ = c/a

(sinθ + cos θ) 2 = sin 2 θ + cos 2 θ + 2 sin θ cos θ,

(b/a) 2 = 1 + 2(c/a) {using the identity sin 2 A + cos 2 A = 1}

b 2 /a 2 = 1 + (2c/a)

b 2 = a 2 + 2ac

a 2 – b 2 + 2ac = 0

2. If tan A = 1/2 and tan B = 1/3, then the value of A + B is

Correct option: (d) π/4

tan A = 1/2, tan B = 1/3

We know that,

tan(A + B) = (tan A + tan B)/(1 – tan A tan B)

Therefore, A + B = π/4

3. The value of cos 1° cos 2° cos 3° … cos 179° is

Correct option: (b) 0

cos 1° cos 2° cos 3° … cos 179°

= cos 1° cos 2° cos 3° … cos 89° cos 90° cos 91° … cos 179°

= cos 1° cos 2° cos 3° … cos 89° (0) cos 91° … cos 179°

= 0 {since the value of cos 90° = 0}

4. The value of sin 50° – sin 70° + sin 10° is equal to

sin 50° – sin 70° + sin 10°

= sin(60° – 10°) – sin(60° + 10°) + sin 10°

Using the formulas

sin(A – B) = sin A cos B – cos A sin B

sin(A + B) = sin A cos B + cos A sin B, we get;

sin 50° – sin 70° + sin 10° = sin 60° cos 10° – cos 60° sin 10° – sin 60° + cos 10° – cos 60° sin 10° + sin 10°

= -2 cos 60° sin 10° + sin 10°

= -2 × (1/2) × sin 10° + sin 10°

= – sin 10° + sin 10°

5. The value of sin (45° + θ) – cos (45° – θ) is

Correct option: (d) 0

sin (45° + θ) – cos (45° – θ)

= sin (45° + θ) – sin (90° -(45° – θ)) {since sin(90° – A) = cos A}

= sin (45° + θ) – sin (45° + θ)

Alternative method:

sin(A + B) = sin A cos B + cos A sin B

cos(A – B) = cos A cos B + sin A sin B, we get;

sin (45° + θ) – cos (45°– θ) = sin 45° cos θ + cos 45° sin θ – cos 45° cos θ – sin 45° sin θ

= (1/√2) cos θ + (1/√2) sin θ – (1/√2) cos θ – (1/√2) sin θ

6. The value of tan 1° tan 2° tan 3° … tan 89° is

(d) Not defined

Correct option: (b) 1

tan 1° tan 2° tan 3° … tan 89°

tan A × cot A =1 and tan 45° = 1

Hence, the equation becomes as;

= 1 × 1 × 1 × 1 × …× 1

= 1 {As 1ⁿ = 1}

7. If α + β = π/4, then the value of (1 + tan α) (1 + tan β) is

Correct option: (b) 2

α + β = π/4

Taking “tan” on both sides,

tan(α + β) = tan π/4

and tan π/4 = 1.

So, (tan α + tan β)/(1 – tan α tan β) = 1

tan α + tan β = 1 – tan α tan β

tan α + tan β + tan α tan β = 1….(i)

(1 + tan α)(1 + tan β) = 1 + tan α + tan β + tan α tan β

8. If A lies in the second quadrant and 3 tan A + 4 = 0, then the value of (2 cot A – 5 cos A + sin A) is equal to

Correct option: (b) 23/10

Given that A lies in the second quadrant and 3 tan A + 4 = 0.

3 tan A = -4

tan A = -4/3

cot A = 1/tan A = -3/4

Using the identity sec 2 A = 1 + tan 2 A,

sec 2 A = 1 + (16/9) = 25/9

sec A = √(25/9)

sec A = -5/3 (in quadrant II secant is negative)

cos A = 1/sec A = -⅗

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

sin A = √(1 – 9/25) = √(16/25) = 4/5 (in quadrant II sine is positive)

2 cot A – 5 cos A + sin A

= 2(-3/4) – 5(-3/5) + (4/5)

= (-3/2) + 3 + (4/5)

= (-15 + 30 + 8)/10

9. If for real values of x, cos θ = x + (1/x), then

(a) θ is an acute angle

(b) θ is right angle

(c) θ is an obtuse angle

(d) No value of θ is possible

Correct option: (d) No value of θ is possible

cos θ = x + (1/x)

cos θ = (x 2 + 1)/x

⇒ x 2 + 1 = x cos θ

⇒ x 2 – x cos θ + 1 = 0

We know that for any real root of the equation ax 2 + bx + c = 0, b 2 – 4ac ≥ 0.

⇒ (-cos θ) 2 – 4 ≥ 0

⇒ cos 2 θ – 4 ≥ 0

⇒ cos 2 θ ≥ 4

⇒ cos θ ≥ ± 2

We know that -1 ≤ cos θ ≤ 1.

Hence, no value of θ is possible.

10. Number of solutions of the equation tan x + sec x = 2 cos x lying in the interval [0, 2π] is

Correct option: (c) 2

tan x + sec x = 2 cos x

(sin x/cos x) + (1/cos x) = 2 cos x

(sin x + 1)/cos x = 2 cos x

sin x + 1 = 2 cos 2 x

sin x + 1 = 2(1 – sin 2 x)

sin x + 1 = 2 – 2 sin 2 x

⇒ 2 sin 2 x + sin x − 1 = 0

⇒ (2 sin x − 1)(sin x + 1)=0

⇒ sin x = −1, 1/2

But for x = 3π/2, tan x and sec x are not defined.

Therefore, there are only two solutions for the given equation in the interval [0, 2π].

Video Lesson on Trigonometry

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