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Class 11 Mathematics Case Study Questions

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If you’re seeking a comprehensive and dependable study resource with Class 11 mathematics case study questions for CBSE, myCBSEguide is the place to be. It has a wide range of study notes, case study questions, previous year question papers, and practice questions to help you ace your examinations. Furthermore, it is routinely updated to bring you up to speed with the newest CBSE syllabus. So, why delay? Begin your path to success with myCBSEguide now!

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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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

Important questions are marked, and Formula sheet is also provided. Click on an exercise or topic to begin.

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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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CBSE Class 11 Term-1 Maths 2021-22: Topic-Wise MCQs, Case-Based & Assertion/Reason

CBSE Class 11 Term-1 Maths 2021-22: Topic-Wise MCQs, Case-Based & Assertion/Reason

In this Post I Have Provided Mathematics  Class 11 Chapter-Wise MCQs, Case-Based Question And Assertion/Reason For Term-1 Session 2021-22 With Solutions. CBSE has Recently Included These Types Of MCQs  And Assertion/Reason For Term-1 Exam 2020.   Every Student Knows these types of Questions Are Very Important For Their Term-1 Examination.

Any student who want to download these MCQs, then they have to click on given respective download links and it will be automatically download in your Google Drive.

CBSE Class 11th  Physics  Term-1 2021-22: Topic-Wise MCQs, Case-Based & Assertion/Reason

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CBSE Sample Papers For Class 11 Mathematics

CBSE sample paper for class  11th Maths  is an important tools to analyse itself. Before going to final examinations every student try to solve different types of sample question paper. It will enhance your knowledge and also provide to increase mental ability. Sample question paper are the best resources for the student to prepare for their Board examinations.

These sample question papers are very helpful to the student to get an entire experience before attempting the Board examinations papers. If students solve different types of sample question papers then they get the good confidence about the appropriate answer and they make good score in their Board examinations.

As every student know CBSE always change their exam pattern and also the model paper for their Board examinations that is why it is very important to all the student to understand the Model paper pattern before going to the final examinations.

In this articles I have provided different types of question papers on the basis of CBSE latest exam pattern which will definitely help to the student for the good preparation for their examinations.

In every year before the Board examination, CBSE provide model test paper to the student, on the basis of these model test papers I have prepared sample question papers for class  11th Maths  which will help to the student to understand the concept of model test paper and also it will help to the student for good preparation for their board examination.

CBSE Sample Paper:

Maths And Physics With Pandey Sir  is a website which provide free study material and notes to every students who are preparing for their Board examinations. By going in Level Section you can select CBSE sample papers. In this section, I have provided CBSE sample paper for class 9th to 12th.

Also if you are preparing for any competitive exams, they can find out study material and important notes in Level Section. In this website I have also provided most important MCQs questions, Assertion/Reason based questions and case study based questions for Board examinations or any other competitive examinations.

Benefits Of Solving CBSE Sample Papers for Examination:

Time Management:-

With the help of CBSE sample question papers you can make a proper strategy for their examinations also by practicing sample question papers you can manage your time during the Board examinations.

Examination Strategy:-

After practicing different types of sample question papers you can understand on which section you have spend more time and which section is easy and which section is hard. In this way you can make a proper examination strategy.

Self Evaluation:-

Self evaluation is very important to every student because with the help of self evaluation, student can understand their weak and strong point, which is very important for any Board examinations. After practicing more and more simple paper you can evaluate himself.

To Identify Silly Mistakes:-

As every student know there are many questions in their question paper which are very easy  but student make mistakes in this types of question. By solving different types of sample question papers you can reduce their silly mistakes.

Advantages Of CBSE Sample Question Papers:

* To get a good exam ideas about your examination paper patterns

* Student can easily access to the question papers

* It is very helpful to know about the mark distribution system.

* Helpful in self evaluation

* To know about the pattern of CBSE

* Help to adjust a proper time management

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Important Questions Class 11 Maths Chapter 3

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Important Questions Class 11 Mathematics Chapter 3

Important questions for cbse class 11 mathematics chapter 3 – trigonometric functions.

Trigonometric functions Class 11 Important Questions have been prepared for Class 11 students to help them score higher in the exams. The important Questions for Trigonometric Functions are prepared by subject matter experts in accordance with the latest CBSE guidelines. 

The Class 11 Mathematics Trigonometric Functions Important Questions include step-by-step solutions for questions ranging from simple to difficult. Furthermore, students can easily access these study materials from Extramarks to have a ready reference to the chapters and questions whenever they need it. Students can also make notes and mark them for quick revision based on the important questions. They can easily develop command over trigonometric functions as they answer these important questions.

CBSE Class 11 Mathematics Chapter-3 Important Questions

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

Students can view the set of important questions given below. 

Q1. Prove that  sin5x−2sin3x+sinx / cos5x−cosx =tanx  

A1. Starting with the left-hand side and using the trigonometric difference identities for the sine function, we obtain

 L.H.S.=sin5x+sinx−2sin3x / cos5x−cosx 

=2sin 3 x.cos 2 x−2sin 3 x /    −2sin 3 x.sin 2 x 

=2sin 3 x(cos 2 x−1)/ −2sin 3 x.sin 2 x 

=−(1−cos 2 x)/−sin 2 x 

=2sin 2 x / 2sinx.cosx 

=sinx / cosx 

Q2. Prove that  cos6x=32cos 2 x − 48cos 4 x + 18cos 2 x − 1  

A2. Starting with the left-hand side and using the trigonometric identities for the cosine function, we obtain

=cos2(3x)=2Cos23x−1 

=2(4cos3x−3cosx)2−1 

=2[16cos6x+9cos2x−24cos4x]−1 

=32cos6x+18cos2x−48cos4x−1 

=32cos6x−48cos4x+18cos2x1 

Q3. Prove that  sin(x+y)/sin(x−y)=tanx+tany/tanx−tany  

A3. Starting with the left-hand side and using the trigonometric difference formula for the sine function, we get

=sin(x+y)/sin(x−y) 

=sinx.cosy+cosx.siny/sinx.cosy−cosx.siny 

Dividing numerator and denominator by  cosx.cosy 

=tanx+tany/tanx−tany 

Q4. The minute hand of a watch is 1.5 cm long. How far does its tip move in 40 minutes? 

A4. Analysing the given information, we have, 

 Angle made in 60min=360 ∘ 

 Angle made in 1min=6 ∘  

 Angle made in 40min=6 ∘ ×40 ∘ =240 ∘  

Calculating the arc distance

240×π/180=l/1.5 

Q6. Show that tan 3x. tan 2x. tan x = tan 3x  tan 2x  tan x  

A6. Let us start with  tan3x  and we know  3x=2x+x 

tan3x=tan(2x+x) 

tan3x/1=tan2x+tanx/1−tan2x.tanx 

tan3x(1−tan2x.tanx)=tan2x+tanx 

tan3x−tan3x.tan2x.tanx=tan2x+tanx 

tan3x.tan2x.tanx=tan3x−tan2x−tanx 

Q7. A wheel makes  360  revolutions in  1  minute. How many radians does it turn in  1  second?

Number of revolutions made in 60s=360 

Number of revolutions made in 1s=360/60 

Angle moved in 6 revolutions = 2π × 6 = 12π 

Given below is the complete set of Important Questions for Trigonometric Functions, which can be accessed by clicking the link provided.

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

In layman’s terms, trigonometric functions are functions of triangle angles. On the basis of these functions, it defines the relationship between the sides and angles of a triangle. The sine, cosine, secant, cosecant, tangent, and cotangent are trigonometric functions. It is also referred to as circular functions. Several trigonometric formulas and identities can be used to define the relationship between angles and functions. 

Tips to Score Marks in Trigonometric Functions

One of the most important chapters in Class 11 Mathematics is the Trigonometric Function. Trigonometry was developed primarily to solve geometric problems involving triangles. Students can easily score high marks in the examinations by practising the important questions of Mathematics Class 11 Trigonometric Functions. When students prepare for these important questions, they can also learn several tricks and shortcuts to answer the questions quickly. Furthermore, students must concentrate on the trigonometric function formulas that are required to solve the sums. Students should not skip this chapter because it covers many important topics, such as designing electronic circuits, calculating tide heights, and so on.

Discuss the Trigonometric Tables and formulas 

The Formula for Function of Trigonometric Ratios

Trigonometric Table

Important Questions for Class 11 Mathematics Chapter 3 Based on Exercise

• In one minute, an engine makes 360 revolutions. How many radians will it turn in a second?

An engine’s total number of revolutions per minute = 360

1 minute equals 60 seconds

As a result, the number of revolutions in one second = 360 / 60 = 6.

360° is the angle formed in one revolution.

Angles are formed in six revolutions, which equals six 360°.

The angle in radians measured over six revolutions = 6 360 / 180

= 6 × 2 × π

As a result, the engine rotates 12 radians in one second. 

Importance of Downloading Class 11 Mathematics Chapter 3 Important Questions

Students will gain a thorough understanding of Trigonometric Functions by accessing the Important Questions for Class 11 Mathematics Chapter 3 . Here are some advantages:

• They can make important notes for the exam.
• They will have access to all Trigonometric Functions Class 11 Important Questions.
• They can use it as a ready reference resource.

It will help  them understand the examination’s question pattern.

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Q.1 If A + B = 225°, then tan A + tan B + tan A × tan B is equal to: 1 0 1 3 Marks :1

Ans tan (A + B) = tan (225) = tan (180+45) = tan 45 = 1

Now, tan(A+B) = (tan A+tan B)/(1 tan A × tan B)

(tan A + tan B)/(1? tan A × tan B) = 1

tan A + tan B + tan A × tan B = 1

Q.2 if ? is an acute angle and sin (?/2) = x-12xthen tan ? is Marks :1

Ans tan?=sin?cos?= 2sin?(?/2)?cos?(?/2)1?sin2(?/2)

tan?=2x?12x?1?x?12×1?2x?12x = x2?1

Q.3 Which is greater ? sin1or sin1? Justify your answer. Marks :4

Ans First, we shall convert 1 into degree µ =180°

1=180 =180227 =180—722 =90—711 =63011 1=57.27

sin1=sin 57.27bHence, sin1 is greater than sin1.

Q.4 If three angles A, B, C, are in A.P. Prove that: cotB=sinA-sinCcosC-cosA Marks :3

Ans R.H.S=sinA-sinCcosC-cosA

=2sinA-C2cosA+C22sinA+C2sinA-C2

=cotA+C2=cotB=L.H.S µA,B,C are in A.P  2B=A+C

Q.5 Show that: 2+2+2+2cos8?=2cos? Marks :4

Ans LH.S=2+2+2(1+cos 8¸) =22+2—2cos2 4 µ 1+cos 8 =2cos2 4 =2+2+4cos2 4 =2+2+2cos 4

=2+21+cos4 =2+2-2cos22µ1+cos4 =2cos22=2+4cos22 =2+2cos2 =21+cos2 =2.2cos2 =2cos=R ·H.S

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Faqs (frequently asked questions), 1. what are the fundamental trigonometric functions.

Trigonometric functions are fundamental functions in mathematics that are used to denote the relationship between the angles of a right-angled triangle and the lengths of its sides.

There are six trigonometric ratios in total, with sine, cosine, and tangent serving as the fundamental three. The other three are called cosecant, secant, and cotangent and are described in relation to previously discussed basic functions.

In a given right-angled triangle, the following values of these ratios are calculated for a specific angle θ in terms of its sides:

sin θ = opposite side / hypotenuse

cos θ = adjacent side / hypotenuse

tan θ = sin θ / cos θ = opposite site / adjacent side

cosec θ = 1 / sin θ

sec θ = 1 / cos θ

cot θ = cos θ / sin θ = 1 / tan θ

2. What is the relationship between Radians and Degrees?

A radian (abbreviated rad or c) is a standard unit of measurement for angles based on the relationship between the length of an arc and the radius of a circle.

rad = arc length / circle radius

Many times, you will be required to convert given angle values  from radians to degrees. The relationship between these two units can be deduced as follows.

The circumference C of a full circle with radius r and angle 360o is equal to the arc length.

Therefore, arc length = C = 2πr.

From the first relation, 360o in radians will be

rad = 2πr / r = 2π

Therefore, 2π rad = 360o,

that is, 1 rad = 180o / π.

This brings us to the final relationship, that is, degree = radian x 180 / π.

3. What are the practical applications of trigonometry?

Trigonometry has a wide range of applications in marine biology, navigation, aviation, and other fields because it involves the relationship between sides and angles, finding the height, and calculating the distance. It is also used to solve complex mathematical problems, such as those in calculus and algebra.

4. What are the angles of trigonometry?

You will come across several new terms that are used in trigonometry questions as you study Chapter 3 Trigonometry of Class 11 Mathematics. The angles of trigonometry are the most important of these. These are essentially functions that aid in the relationship of triangle sides and angles. The sine, cosine, and tangent are abbreviated as sin, cos, and tan. Secant, cosecant, and cotangent are the inverse functions or angles, abbreviated as sec, cosec, and cot.

5. Name the Six Trigonometric Functions.

The six trigonometric functions are sine, cosine, tan, cosec, sec, and cot.

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Trigonometric Functions Class 11 Notes: CBSE Maths Chapter 3

• Revision Notes
• Chapter 3 Trigonometric Functions

Class 11 Maths Chapter 3 Trigonometric Functions Notes FREE PDF Download

Trigonometry is a crucial part of Class 11 Mathematics, and mastering it is essential for success in both board exams and competitive exams. In Chapter 3, Trigonometric Functions, you’ll explore the fundamental concepts, identities, and applications of trigonometric functions. Our comprehensive notes simplify these complex topics, providing clear explanations, important formulas, and solved examples to help you easily grasp the subject.

Take advantage of the FREE PDF download to access these valuable resources anytime, anywhere. Visit our pages to get your class 11 Maths notes and check out the Class 11 Maths syllabus to stay on track with your studies.

Access Revision Notes for Class 11 Maths Chapter 3 Trigonometric Functions

1. the meaning of trigonometry.

${{\text{Tri }}}{{\text{ Gon }}}{{\text{ Metron }}}$

$\downarrow\quad\;\;\;\;\downarrow\quad\;\;\;\;\downarrow$

$3\quad{{\text{ sides }}}{{\text{ Measure }}}$

As a result, this area of mathematics was established in the ancient past to measure a triangle's three sides, three angles, and six components. Time-trigonometric functions are utilised in a variety of ways nowadays. The sine and cosine of an angle in a right-angled triangle are the two fundamental functions, and there are four more derivative functions.

2. Basic Trigonometric Identities

(a) ${\sin ^2}\theta  + {\cos ^2}\theta  = 1: - 1 \leqslant \sin \theta  \leqslant 1; - 1 \leqslant \cos \theta  \leqslant 1\forall \theta  \in {\text{R}}$

(b) ${\sec ^2}\theta  - {\tan ^2}\theta  = 1:|\sec \theta | \geqslant 1\forall \theta  \in {\text{R}}$

(c) ${\operatorname{cosec} ^2}\theta  - {\cot ^2}\theta  = 1:|\operatorname{cosec} \theta | \geqslant 1\forall \theta  \in {\text{R}}$

Trigonometric Ratios of Standard Angles:

The relation between these trigonometric identities with the sides of the triangles can be given as follows:

Sine θ $=$ Opposite/Hypotenuse

Cos θ  $=$ Adjacent/Hypotenuse

Tan θ  $=$ Opposite/Adjacent

Cot θ $=$ Adjacent/Opposite

Cosec θ  = Hypotenuse/Opposite

Sec θ  = Hypotenuse/Adjacent

The following are the signs of trigonometric ratios in different quadrants:

3. Trigonometric Ratios of Allied Angles

We might calculate the trigonometric ratios of angles of any value using the trigonometric ratio of allied angles.

1. Sin(–θ)=–Sinθ

2. Cos(–θ)=Cosθ

3. Tan(–θ)=–Tanθ

4. Sin(90 o –θ)=Cosθ

5. Cos(90 o –θ)=Sinθ

6. Tan(90 o –θ)=Cotθ

7. Sin(180 o –θ)=Sinθ

8. Cos(180 o –θ)=–Cosθ

9. Tan(180 o –θ)=–Tanθ

10. Sin(270 o –θ)=–Cosθ

11. Cos(270 o –θ)=–Sinθ

12. Tan(270 o –θ)=Cotθ

13. Sin(90 o +θ)=Cosθ

14. Cos(90 o +θ)=–Sinθ

15. Tan(90 o +θ)=–Cotθ

16. Sin(180 o +θ)=–Sinθ

17. Cos(180 o +θ)=–Cosθ

18. Tan(180 o +θ)=Tanθ

19. Sin(270 o +θ)=–Cosθ

20. Cos(270 o +θ)=Sinθ

21. Tan(270 o +θ)=–Cotθ

4. Trigonometric Functions of Sum or Difference of Two Angles

(a) $\sin ({\text{A}} + {\text{B}}) = \sin {\text{A}}\cos {\text{B}} + \cos {\text{A}}\sin {\text{B}}$

(b) $\sin ({\text{A}} - {\text{B}}) = \sin {\text{A}}\cos {\text{B}} - \cos {\text{A}}\sin {\text{B}}$

(c) $\cos (A + B) = \cos A\cos B - \sin A\sin B$

(d) $\cos ({\text{A}} - {\text{B}}) = \cos {\text{A}}\cos {\text{B}} + \sin {\text{A}}\sin {\text{B}}$

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

(f) $\tan ({\text{A}} - {\text{B}}) = \dfrac{{\tan {\text{A}} - \tan {\text{B}}}}{{1 + \tan {\text{A}}\tan {\text{B}}}}$

(g) $\cot (A + B) = \dfrac{{\cot A\cot B - 1}}{{\cot B + \cot A}}$

(f) $\cot ({\text{A}} - {\text{B}}) = \dfrac{{\cot {\text{A}}\cot {\text{B}} + 1}}{{\cot {\text{B}} - \cot {\text{A}}}}$

(h) ${\sin ^2}\;{\text{A}} - {\sin ^2}\;{\text{B}} = {\cos ^2}\;{\text{B}} - {\cos ^2}\;{\text{A}} = \sin ({\text{A}} + {\text{B}}) \cdot \sin ({\text{A}} - {\text{B}})$

(i) ${\cos ^2}\;{\text{A}} - {\sin ^2}\;{\text{B}} = {\cos ^2}\;{\text{B}} - {\sin ^2}\;{\text{A}} = \cos ({\text{A}} + {\text{B}}) \cdot \cos ({\text{A}} - {\text{B}})$

(j) $\tan (A + B + C) = \dfrac{{\tan A + \tan B + \tan C - \tan A\tan B\tan C}}{{1 - \tan A\tan B - \tan B\tan C - \tan C\tan A}}$

5. Multiple Angles and Half Angles

(a) $\sin 2\;{\text{A}} = 2\sin {\text{A}}\cos {\text{A}};\quad \sin \theta  = 2\sin \dfrac{\theta }{2}\cos \dfrac{\theta }{2}$

(b) $\cos 2\;{\text{A}} = {\cos ^2}\;{\text{A}} - {\sin ^2}\;{\text{A}} = 2{\cos ^2}\;{\text{A}} - 1 = 1 - 2{\sin ^2}\;{\text{A}}$  

$2{\cos ^2}\dfrac{\theta }{2} = 1 + \cos \theta ,2{\sin ^2}\dfrac{\theta }{2} = 1 - \cos \theta $

(c) $\tan 2\;{\text{A}} = \dfrac{{2\tan {\text{A}}}}{{1 - {{\tan }^2}\;{\text{A}}}};\tan \theta  = \dfrac{{2\tan \dfrac{\theta }{2}}}{{1 - {{\tan }^2}\dfrac{\theta }{2}}}$

(d) $\sin 2\;{\text{A}} = \dfrac{{2\tan {\text{A}}}}{{1 - {{\tan }^2}\;{\text{A}}}};\cos 2\;{\text{A}} = \dfrac{{1 - {{\tan }^2}\;{\text{A}}}}{{1 + {{\tan }^2}\;{\text{A}}}}$

(e) $\sin 3\;{\text{A}} = 3\sin {\text{A}} - 4{\sin ^3}\;{\text{A}}$

(f) $\cos 3\;{\text{A}} = 4{\cos ^3}\;{\text{A}} - 3\cos {\text{A}}$

(g) $\tan 3\;{\text{A}} = \dfrac{{3\tan {\text{A}} - {{\tan }^3}\;{\text{A}}}}{{1 - 3{{\tan }^2}\;{\text{A}}}}$

6. Transformation of Products into Sum or Difference of Sines & Cosines

(a) $2\sin {\text{A}}\cos {\text{B}} = \sin ({\text{A}} + {\text{B}}) + \sin ({\text{A}} - {\text{B}})$

(b) $2\cos {\text{A}}\sin {\text{B}} = \sin ({\text{A}} + {\text{B}}) - \sin ({\text{A}} - {\text{B}})$

(c) $2\cos {\text{A}}\cos {\text{B}} = \cos ({\text{A}} + {\text{B}}) + \cos ({\text{A}} - {\text{B}})$

(d) $2\sin {\text{A}}\sin {\text{B}} = \cos ({\text{A}} - {\text{B}}) - \cos ({\text{A}} + {\text{B}})$

7. Factorisation of the Sum or Difference of Two Sines or Cosines

(a) $\sin {\text{C}} + \sin {\text{D}} = 2\sin \dfrac{{{\text{C}} + {\text{D}}}}{2}\cos \dfrac{{{\text{C}} - {\text{D}}}}{2}$

(b) $\sin {\text{C}} - \sin {\text{D}} = 2\cos \dfrac{{{\text{C}} + {\text{D}}}}{2}\sin \dfrac{{{\text{C}} - {\text{D}}}}{2}$

(c) $\cos C + \cos D = 2\cos \dfrac{{C + D}}{2}\cos \dfrac{{C - D}}{2}$

(d) $\cos {\text{C}} - \cos {\text{D}} =  - 2\sin \dfrac{{{\text{C}} + {\text{D}}}}{2}\sin \dfrac{{{\text{C}} - {\text{D}}}}{2}$

8. Important Trigonometric Ratios

(a) $\sin {\text{n}}\pi  = 0;\cos {\text{n}}\pi  = {( - 1)^{\text{n}}};\tan {\text{n}}\pi  = 0$ where ${\text{n}} \in {\text{Z}}$

(b) $\sin {15^\circ }$ or $\sin \dfrac{\pi }{{12}} = \dfrac{{\sqrt 3  - 1}}{{2\sqrt 2 }} = \cos {75^\circ }$ or $\cos \dfrac{{5\pi }}{{12}}$ ; 

$\cos {15^\circ }$ or $\cos \dfrac{\pi }{{12}} = \dfrac{{\sqrt 3  + 1}}{{2\sqrt 2 }} = \sin {75^\circ }$ or $\sin \dfrac{{5\pi }}{{12}}$  

$\tan {15^\circ } = \dfrac{{\sqrt 3  - 1}}{{\sqrt 3  + 1}} = 2 - \sqrt 3  = \cot {75^\circ }$  

$\tan {75^\circ } = \dfrac{{\sqrt 3  + 1}}{{\sqrt 3  - 1}} = 2 + \sqrt 3  = \cot {15^\circ }$

(c) $\sin \dfrac{\pi }{{10}}$ or $\sin {18^\circ } = \dfrac{{\sqrt 5  - 1}}{4}$ & $\cos {36^\circ }$ or $\cos \dfrac{\pi }{5} = \dfrac{{\sqrt 5  + 1}}{4}$

9. Conditional Identities

If ${\text{A}} + {\text{B}} + {\text{C}} = \pi $ then :

(i) $\sin 2\;{\text{A}} + \sin 2\;{\text{B}} + \sin 2{\text{C}} = 4\sin {\text{A}}\sin {\text{B}}\sin {\text{C}}$

(ii) $\sin {\text{A}} + \sin {\text{B}} + \sin {\text{C}} = 4\cos \dfrac{{\text{A}}}{2}\cos \dfrac{{\text{B}}}{2}\cos \dfrac{{\text{C}}}{2}$

(iii) $\cos 2\;{\text{A}} + \cos 2\;{\text{B}} + \cos 2{\text{C}} =  - 1 - 4\cos {\text{A}}\cos {\text{B}}\cos {\text{C}}$

(iv) $\cos {\text{A}} + \cos {\text{B}} + \cos {\text{C}} = 1 + 4\sin \dfrac{{\text{A}}}{2}\sin \dfrac{{\text{B}}}{2}\sin \dfrac{{\text{C}}}{2}$

(v) $\tan {\text{A}} + \tan {\text{B}} + \tan {\text{C}} = \tan {\text{A}}\tan {\text{B}}\tan {\text{C}}$

(vi) $\tan \dfrac{{\text{A}}}{2}\tan \dfrac{{\text{B}}}{2} + \tan \dfrac{{\text{B}}}{2}\tan \dfrac{{\text{C}}}{2} + \tan \dfrac{{\text{C}}}{2}\tan \dfrac{{\text{A}}}{2} = 1$

(vii) $\cot \dfrac{{\text{A}}}{2} + \cot \dfrac{{\text{B}}}{2} + \cot \dfrac{{\text{C}}}{2} = \cot \dfrac{{\text{A}}}{2} \cdot \cot \dfrac{{\text{B}}}{2} \cdot \cot \dfrac{{\text{C}}}{2}$

(viii) $\cot A\cot B + \cot B\cot C + \cot C\cot A = 1$

10. Range of Trigonometric Expression

${{\text{E}} = {\text{a}}\sin \theta  + {\text{b}}\cos \theta }$

${{\text{E}} = \sqrt {{{\text{a}}^2} + {{\text{b}}^2}} \sin (\theta  + \alpha ),\left( {{\text{ where }}\tan \alpha  = \dfrac{{\text{b}}}{{\text{a}}}} \right)}$ 

${{\text{E}} = \sqrt {{{\text{a}}^2} + {{\text{b}}^2}} \cos (\theta  - \beta ),\left( {{\text{ where }}\tan \beta  = \dfrac{{\text{a}}}{{\text{b}}}} \right)}$

Hence for any real value of $\theta , - \sqrt {{a^2} + {b^2}}  \leqslant E \leqslant \sqrt {{a^2} + {b^2}} $

The trigonometric functions are very important for studying triangles, light, sound or waves. The values of these trigonometric functions in different domains and ranges can be used from the following table:

Trigonometric Functions in Different Domains and Ranges

11. Sine and Cosine Series

(a) $\quad \sin \alpha  + \sin (\alpha  + \beta ) + \sin (\alpha  + 2\beta ) +  \ldots . + \sin (\alpha  + \overline {n - 1} \beta )$

$ = \dfrac{{\sin \dfrac{{{\text{n}}\beta }}{2}}}{{\sin \dfrac{\beta }{2}}}\sin \left( {\alpha  + \dfrac{{{\text{n}} - 1}}{2}\beta } \right)$

${\cos \alpha  + \cos (\alpha  + \beta ) + \cos (\alpha  + 2\beta ) +  \ldots  + \cos (\alpha  + \overline {n - 1} \beta )}$

${ = \dfrac{{\sin \dfrac{{n\beta }}{2}}}{{\sin \dfrac{\beta }{2}}}\cos \left( {\alpha  + \dfrac{{n - 1}}{2}\beta } \right)}$ 

12. Graphs of Trigonometric Functions 

(a). ${y = \sin x,}$

${x \in R;y \in [ - 1,1]}$ 

(b). $y = \cos x$

$x \in R;y \in [ - 1,1]$ 

(c) ${y = \tan x}$

${x \in R - \left\{ {(2n + 1)\dfrac{\pi }{2};n \in Z} \right\};y \in R}$ 

(d) ${y = \cot x}$

${x \in R - \{ n\pi ;n \in z\} ;y \in R}$ 

(e) ${y = \operatorname{cosec} x}$ 

${x \in R - \{ n\pi ;n \in Z\} ;y \in ( - \infty , - 1] \cup [1,\infty )}$

(f) $y = \sec x\quad $

$x \in R - \left\{ {(2n + 1)\dfrac{\pi }{2};n \in Z} \right\};y \in ( - \infty , - 1] \cup [1,\infty )$ 

Trigonometric Equations

13. Trigonometric Equations

Trigonometric equations are equations using trigonometric functions with unknown angles. 

e.g.,  $\cos \theta  = 0,{\cos ^2}\theta  - 4\cos \theta  = 1$ .

The value of the unknown angle that satisfies a trigonometric equation is called a solution.

e.g., $\quad \sin \theta  = \dfrac{1}{{\sqrt 2 }} \Rightarrow \theta  = \dfrac{\pi }{4}$ or $\theta  = \dfrac{\pi }{4},\dfrac{{3\pi }}{4},\dfrac{{9\pi }}{4},\dfrac{{11\pi }}{4}, \ldots $

As a result, the trigonometric equation can have an unlimited number of solutions and is categorised as follows:

(i). Principal Solution

As we know, the values of $\sin x$ and $\cos x$ will get repeated after an interval of $2 \pi$. In the same way, the values of $\tan x$ will get repeated after an interval of $\pi$. 

So, if the equation has a variable $0 \leq \mathrm{x}<2 \pi$, then the solutions will be termed as principal solutions. 

Find the principal solutions of the equation $\sin x=\dfrac{\sqrt{3}}{2}$.

Solution: We know that, $\sin \dfrac{\pi}{3}=\dfrac{\sqrt{3}}{2}$

Also, $\sin \dfrac{2 \pi}{3}=\sin \left(\pi-\dfrac{\pi}{3}\right)$

Now, we know that $\sin (\pi-x)=\sin x$. 

Hence, $\sin \dfrac{2 \pi}{3}=\sin \dfrac{\pi}{3}=\dfrac{\sqrt{3}}{2}$

Therefore, the principal solutions of $\sin x=\dfrac{\sqrt{3}}{2}$ are $\mathrm{x}=\dfrac{\pi}{3}$ and $\dfrac{2 \pi}{3}$.

(ii). General solution

A general solution is one that involves the integer 'n' and yields all trigonometric equation solutions. Also, the character ' $\mathrm{Z}$ ' is used to denote the set of integers.

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

Solution: We know that $\sin \dfrac{\pi}{3}=\dfrac{\sqrt{3}}{2}$. Therefore, $\sin x=-\dfrac{\sqrt{3}}{2}=-\sin \dfrac{\pi}{3}$

Using the unit circle properties, we get $\sin x=-\sin \dfrac{\pi}{3}=\sin \left(\pi+\dfrac{\pi}{3}\right)=\sin \dfrac{4 \pi}{3}$ Hence, $\sin x=\sin \dfrac{4 \pi}{3}$

Since, we know that for any real numbers $x$ and $y, \sin x=\sin y$ implies $x=n \pi+(-1)^{n} y$, where $n \in Z$.

So, we get, $x=n \pi+(-1)^{\mathrm{n}}\left(\dfrac{4 \pi}{3}\right)$

14. Results

1. $\quad \sin \theta  = 0 \Leftrightarrow \theta  = \operatorname{n} \pi $

2. $\cos \theta  = 0 \Leftrightarrow \theta (2{\text{n}} + 1)\dfrac{\pi }{2}$

3. $\tan \theta  = 0 \Leftrightarrow \theta  = {\text{n}}\pi $

4. $\sin \theta  = \sin \alpha  \Leftrightarrow \theta  = {\text{n}}\pi  + {( - 1)^{\text{n}}}\alpha $ , where $\alpha  \in \left[ { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right]$

5. $\cos \theta  = \cos \alpha  \Leftrightarrow \theta  = 2{\text{n}}\pi  \pm \alpha $ , where $\alpha  \in [0,\pi ]$

6. $\tan \theta  = \tan \alpha  \Leftrightarrow \theta  = {\text{n}}\pi  + \alpha $ , where $\alpha  \in \left( { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right)$

7. ${\sin ^2}\theta  = {\sin ^2}\alpha  \Leftrightarrow \theta  = {\text{n}}\pi  \pm \alpha $ .

8. ${\cos ^2}\theta  = {\cos ^2}\alpha  \Leftrightarrow \theta  = $ n $\pi  \pm \alpha $ .

9. ${\tan ^2}\theta  = {\tan ^2}\alpha  \Leftrightarrow \theta  = {\text{n}}\pi  \pm \alpha $ .

10. $\sin \theta  = 1 \Leftrightarrow \theta  = (4{\text{n}} + 1)\dfrac{\pi }{2}$

11. $\cos \theta  = 1 \Leftrightarrow \theta  = 2{\text{n}}\pi $

12. $\cos \theta  =  - 1 \Leftrightarrow \theta  = (2{\text{n}} + 1)\pi $ .

13. $\sin \theta  = \sin \alpha $ and $\cos \theta  = \cos \alpha  \Leftrightarrow \theta  = 2{\text{n}}\pi  + \alpha $

Steps to Solve Trigonometric Functions:

The following are the stages of solving trigonometric equations:

Step 1: Decompose the trigonometric equation into a single trigonometric ratio, preferably the sine or cos function.

Step 2: Factor the trigonometric polynomial given in terms of the ratio.

Step 3: Write down the general solution after solving for each factor.

1. Unless otherwise stated, is treated as an integer throughout this chapter.

2. Unless the answer is required in a specific interval or range, the general solution should be supplied.

3. The angle's main value is regarded as $\alpha $ . (The main value is the angle with the least numerical value.)

Trigonometric Functions Class 11 Notes Maths - Basic Subjective Questions

Section–a (1 mark questions).

1. If $tan\Theta =\frac{-4}{3}$ , then find $sin\Theta$.

Ans . Since, $\tan \theta=-\frac{4}{3}$ is negative, $\theta$ lies either in second quadrant or in fourth quadrant.

$$ \begin{aligned} & \because 1+\cot ^2 \theta=\operatorname{cosec}^2 \theta \\ & \Rightarrow 1+\frac{1}{\left(-\frac{4}{3}\right)^2}=\frac{1}{\sin ^2 \theta} \\ & \Rightarrow \sin ^2 \theta=\left(\frac{4}{5}\right)^2 \end{aligned} $$

Thus $\sin \theta=\frac{4}{5}$ if $\theta$ lies in the second quadrant or $\sin \theta=-\frac{4}{5}$, if $\theta$ lies in the fourth quadrant.

2. Find the greatest value of sin x cos x .

Ans. $\sin x \cos x=\frac{1}{2}(2 \sin x \cos x)=\frac{1}{2} \sin 2 x$ Greatest value of $\sin 2 x=1$

$\therefore$ Greatest value of $\dfrac{1}{2} \sin 2 x=\dfrac{1}{2} \times 1=\dfrac{1}{2}$

3. Find the degree measure of angle $\left ( \frac{\pi }{8} \right )^{c}$  radian.

Ans. We know that, $\pi$ radians $=180^{\circ}$

$$ \begin{aligned} & \Rightarrow \frac{\pi}{8} \text { radians }=22.5^{\circ} \\ & =22^{\circ}+0.5^{\circ}=22^{\circ}+30^{\prime}=22^{\circ} 30^{\prime} \end{aligned} $$

4. Evaluate: $sin \left ( \frac{-15\pi }{4} \right )$ .

$$ \begin{aligned} & \text { 4. } \sin \left(\frac{-15 \pi}{4}\right)=\sin \left(\frac{-16 \pi+\pi}{4}\right) \\ & =\sin \left(-4 \pi+\frac{\pi}{4}\right) \\ & =\sin \left(\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}} \end{aligned} $$

5. The minute hand of a watch is 1.5 cm long. How far does it tip move in 40 minutes? (Use $\pi$ = 3.14) .

Ans. Angle rotated by minute hand in 60 minutes = $2\pi$     radians 

Therefore, angle rotated by minute hand in 40 minutes = $\frac{40}{60}\times2\pi =\frac{4\pi }{3}$  radians.

Hence, the required distance travelled is given by

$l=r\Theta =1.5\times\frac{4\pi }{3}cm=2\pi\;cm$

$=2\times3.14\;cm=6.28\;cm$

Section–B (2 Marks Questions)

6. Prove that cot2x cotx - cot3x cot2x - cot3x cotx = 1

Ans. We have, cot 3x=cot (2x+x)

$\Rightarrow cot\;3x=\dfrac{cot\;2x\;cot\;x -1}{cot\;2x+cot\;x}$

$\Rightarrow$ Cot 3x cot 2x + cot 3x cot x = cot 2x cot x-1

$\Rightarrow$ cot 2x cot x - cot 3x cot 2x - cot 3x xot x=1

7. Evaluate: $\frac{1-\tan^2{15^\circ}}{1+\tan^2{15^\circ}}$ 

Ans. Given that: $\frac{1-\tan^2{15^\circ}}{1+\tan^2{15^\circ}}$

Let $\Theta =15^{\circ}$

$\therefore 2\Theta =30^{\circ}$

We know that, $\therefore cos\;2\Theta =\frac{1-\tan^2{15^\circ}}{1+\tan^2{15^\circ}}$

$\Rightarrow  cos\;30 =\frac{1-\tan^2{15^\circ}}{1+\tan^2{15^\circ}}$

$\frac{1-\tan^2{15^\circ}}{1+\tan^2{15^\circ}}=\frac{\sqrt{3}}{3}$

8. If for real value of x , $cos\Theta =x+\frac{1}{x}$ then what can you say about $\Theta$    ?

Ans. Given that: $cos\Theta =x+\frac{1}{x}$ 

$\Rightarrow cos\Theta =x+\frac{1}{x}$

$$ \begin{aligned} & \Rightarrow x^2+1=x \cos \theta \\ & \Rightarrow x^2-x \cos \theta+1=0 \end{aligned} $$

For real value of $x, b^2-4 a c \geq 0$

$$ \begin{aligned} & \Rightarrow(-\cos \theta)^2-4 \times 1 \times 1 \geq 0 \\ & \Rightarrow \cos ^2 \theta-4 \geq 0 \\ & \Rightarrow \cos ^2 \theta \geq 4 \\ & \Rightarrow \cos \theta \geq \pm 2 \quad[\because-1 \leq \cos \theta \leq 1] \end{aligned} $$

So, the value of $\theta$ is not possible.

9. Find the value of $sin\left ( \frac{\pi }{4}+\Theta  \right )-cos \left ( \frac{\pi }{4}-\Theta  \right )$ .

Ans. Given expression :

$$ \begin{aligned} & \sin \left(\frac{\pi}{4}+\theta\right)-\cos \left(\frac{\pi}{4}-\theta\right) \\ & \sin \left(\frac{\pi}{4}+\theta\right)=\sin \left(\frac{\pi}{4}\right) \cos \theta+\cos \left(\frac{\pi}{4}\right) \sin \theta \\ & =\frac{1}{\sqrt{2}} \cos \theta+\frac{1}{\sqrt{2}} \sin \theta \\ & \cos \left(\frac{\pi}{4}-\theta\right)=\cos \left(\frac{\pi}{4}\right) \cos \theta+\sin \left(\frac{\pi}{4}\right) \sin \theta \\ & =\frac{1}{\sqrt{2}} \cos \theta+\frac{1}{\sqrt{2}} \sin \theta \\ & \therefore \sin \left(\frac{\pi}{4}+\theta\right)-\cos \left(\frac{\pi}{4}-\theta\right) \\ & =\frac{1}{\sqrt{2}} \cos \theta+\frac{1}{\sqrt{2}} \sin \theta-\frac{1}{\sqrt{2}} \cos \theta-\frac{1}{\sqrt{2}} \sin \theta \\ & =0 \end{aligned} $$

10. Find the value of $tan\;\frac{\pi }{12}$ .

Ans. $\tan \frac{\pi}{12}=\tan \left(\frac{\pi}{4}-\frac{\pi}{6}\right)$

$$ \begin{aligned} & =\frac{\tan \frac{\pi}{4}-\tan \frac{\pi}{6}}{1+\tan \frac{\pi}{4} \cdot \tan \frac{\pi}{6}}=\frac{1-\frac{1}{\sqrt{3}}}{1+1 \cdot \frac{1}{\sqrt{3}}}=\frac{\sqrt{3}-1}{\sqrt{3}+1} \\ & =\frac{(\sqrt{3}-1)(\sqrt{3}-1)}{(\sqrt{3}+1)(\sqrt{3}-1)}=\frac{4-2 \sqrt{3}}{2}=2-\sqrt{3} \end{aligned} $$

11. Prove that : $tan\;225^{\circ}\;cot\;405^{\circ}+tan\;765^{\circ}\;cot\;675^{\circ}=0$ 

$$ \begin{aligned} & \text { 11. LHS }= \\ & \tan 225^{\circ} \cot 405^{\circ}+\tan 765^{\circ} \cot 675^{\circ} \\ & =\tan \left(180^{\circ}+45^{\circ}\right) \cot \left(360^{\circ}+45^{\circ}\right) \\ & \quad \quad \quad \quad \tan \left(360^{\circ} \times 2+45^{\circ}\right) \cot \left(360^{\circ} \times 2-45^{\circ}\right) \\ & =\tan 45^{\circ} \cot 45^{\circ}+\tan 45^{\circ}\left[-\cot 45^{\circ}\right] \\ & =1 \times 1-1 \times 1 \\ & =0=\text { RHS } \end{aligned} $$

Hence proved.

12. For $0<x<\frac{\pi }{2}$ , show that $\sqrt{\frac{(1-cos2x)}{1+cos2x}}=tan\;x$ 

$$ \begin{aligned} & \text { LHS }=\sqrt{\frac{(1-\cos 2 x)}{(1+\cos 2 x)}} \\ & =\sqrt{\frac{1-\left(1-2 \sin ^2 x\right)}{1+\left(2 \cos ^2 x-1\right)}}=\sqrt{\frac{2 \sin ^2 x}{2 \cos ^2 x}} \\ & =|\tan x|=\tan x \quad\left\{\because 0<x<\frac{\pi}{2}\right\} \\ & =\text { RHS } \end{aligned} $$

13. Prove that: $\frac{cos(2\pi +x)cosec(2\pi +x)tan\left ( \frac{\pi }{2}+x \right )}{sec\left ( \frac{\pi }{2}+x \right )cos\;x\;cot(\pi +x)}=1$ 

Ans. LHS $=$

$$\cos (2 \pi+x) \operatorname{cosec}(2 \pi+x) \tan \left(\frac{\pi}{2}+x\right)$$

$$ \begin{aligned} & \sec \left(\frac{\pi}{2}+x\right) \cos x \cot (\pi+x) \\ \Rightarrow & \frac{-\cos x \operatorname{cosec} x \cot x}{-\operatorname{cosec} x \cos x \cot x}=1 \end{aligned} $$

Important Formulas of Class 11 Chapter 3 you shouldn’t Miss!

Basic Trigonometric Identities :

$\sin^2 \theta + \cos^2 \theta = 1$

$1 + \tan^2 \theta = \sec^2 \theta$

$1 + \cot^2 \theta = \csc^2 \theta$

Angle Sum and Difference Formulas :

$\sin(A \pm B) = \sin A \cos B \pm \cos A \sin B$

$\cos(A \pm B) = \cos A \cos B \mp \sin A \sin B$

$\tan(A \pm B) = \frac{\tan A \pm \tan B}{1 \mp \tan A \tan B}$

Double Angle Formulas :

$\sin 2\theta = 2 \sin \theta \cos \theta$

$\cos 2\theta = \cos^2 \theta - \sin^2 \theta$

$\tan 2\theta = \frac{2 \tan \theta}{1 - \tan^2 \theta}$

Product-to-Sum Formulas :

$\sin A \sin B = \frac{1}{2} [\cos(A - B) - \cos(A + B)]$

$\cos A \cos B = \frac{1}{2} [\cos(A - B) + \cos(A + B)]$

$\sin A \cos B = \frac{1}{2} [\sin(A + B) + \sin(A - B)]$

Inverse Trigonometric Functions :

$\sin^{-1} (\sin \theta) = \theta, \quad \text{for} \ -\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}

$\cos^{-1} (\cos \theta) = \theta, \quad \text{for} \ 0 \leq \theta \leq \pi

$\tan^{-1} (\tan \theta) = \theta, \quad \text{for} \ -\frac{\pi}{2} < \theta < \frac{\pi}{2}$

Importance of Trigonometric Functions Class 11 Notes

The importance of Trigonometric Function Class 11 notes lies in their role in helping students build a strong foundation in trigonometry, which is essential for advanced mathematical studies. Here’s why these notes are crucial:

Clear Conceptual Understanding: The notes simplify complex trigonometric concepts, making it easier for students to grasp fundamental ideas like identities, functions, and equations.

Exam Preparation: Well-structured notes provide a comprehensive review of key formulas, theorems, and problem-solving techniques, which are vital for scoring well in exams.

Application in Higher Studies: Trigonometry forms the basis for various topics in Class 12 and competitive exams like JEE. Mastery of this chapter ensures a smoother transition to advanced topics.

Problem-solving Skills: These notes offer step-by-step solutions to typical problems, helping students develop strong analytical and problem-solving abilities.

Quick Revision: Concise and organised, these notes are perfect for quick revision before exams, ensuring students can recall essential concepts and formulas easily.

Tips for Learning the Class 11 Maths Chapter 3 Trigonometric Functions

Here are some helpful tips for learning the Class 11 Maths Chapter 3 Trigonometric Functions:

Master the Basics : Ensure you have a strong understanding of basic trigonometric ratios and identities from earlier classes. This foundation is crucial for tackling more complex concepts in this chapter.

Memorise Key Formulas : Focus on memorising important trigonometric identities, angle sum and difference formulas, and double angle formulas. These are frequently used in solving problems and are essential for success.

Practice Regularly : Trigonometry requires consistent practice. Work through a variety of problems from your textbook and additional reference materials to become comfortable with different types of questions.

Visualise with Unit Circle : Use the unit circle to visualise the relationships between different trigonometric functions. This can help in understanding how angles and functions are related.

Work on Inverse Functions : Spend extra time understanding inverse trigonometric functions, as they can be tricky. Practice converting between trigonometric functions and their inverses.

Mastering the concepts in Class 11 Maths Chapter 3, Trigonometric Functions, is essential for building a solid foundation in mathematics. By focusing on understanding key formulas, practising regularly, and applying the tips provided, students can confidently tackle trigonometric problems and excel in their exams. Regular revision and consistent practice are key to success in this crucial chapter.

Related Study Materials for Class 11 Maths Chapter 3 Trigonometric Functions

Students can also download additional study materials provided by Vedantu for 

Class 11 Maths Trigonometric Functions.

Revision Notes Links for Class 11 Maths

Important Study Materials for Class 11 Maths

FAQs on Trigonometric Functions Class 11 Notes: CBSE Maths Chapter 3

1. What are the key topics covered in the trigonometric functions Class 11 notes?

The trigonometric functions Class 11 notes cover essential topics such as trigonometric identities, angle sum and difference formulas, double angle formulas, and inverse trigonometric functions.

2. How can the trigonometry Class 11 notes help in exam preparation?

Trigonometry Class 11 notes provide clear explanations of concepts, important formulas, and solved examples, making it easier for students to understand and revise the material, thereby enhancing their exam preparation.

3. Are the trigonometric functions Class 11 PDF notes sufficient for board exam preparation?

Yes, trigonometric functions Class 11 PDF notes are designed to cover the entire syllabus comprehensively, making them an excellent resource for board exam preparation.

4. What is the importance of studying trigonometric functions in Class 11?

Studying trigonometric functions in Class 11 is crucial as it lays the groundwork for more advanced topics in Class 12 and competitive exams, and helps in developing problem-solving skills in mathematics.

5. How can trigonometry Class 11 notes PDF be used for quick revision?

Trigonometry Class 11 notes PDF are concise and well-organized, making them perfect for quick revision before exams. Students can easily review key formulas and concepts.

6. Do trigonometric functions Class 11 notes include solved examples?

Yes, trigonometric functions Class 11 notes typically include solved examples that demonstrate how to apply formulas and concepts to solve various types of problems.

7. Can I use trigonometry Class 11 notes for competitive exam preparation?

The trigonometry Class 11 notes are aligned with the syllabus of many competitive exams, providing a strong foundation for solving trigonometry-related questions in those exams.

8. How often should I revise using trigonometric functions Class 11 PDF notes?

Regular revision using trigonometric functions Class 11 PDF notes is recommended to keep the concepts and formulas fresh in your memory, especially as you approach exams.

9. Where can I download high-quality trigonometric functions Class 11 PDF notes?

High-quality trigonometric functions Class 11 PDF notes can be downloaded from reputable educational platforms like Vedantu, which provide comprehensive study materials for free.

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Download Case Study Questions for Class 11 Maths

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[PDF] Download Case Study Questions for Class 11 Maths

Here we are providing case study questions for Class 11 Maths. In this article, we are sharing links for Class 11 Maths All Chapters. All case study questions of Class 11 Maths are solved so that students can check their solutions after attempting questions.

Click on the chapter to view.

Class 11 Maths Chapters

Chapter 1 Sets Chapter 2 Relations and Functions Chapter 3 Trigonometric Functions Chapter 4 Principle of Mathematical Induction Chapter 5 Complex Numbers and Quadratic Equations Chapter 6 Linear Inequalities Chapter 7 Permutation and Combinations Chapter 8 Binomial Theorem Chapter 9 Sequences and Series Chapter 10 Straight Lines Chapter 11 Conic Sections Chapter 12 Introduction to Three-Dimensional Geometry Chapter 13 Limits and Derivatives Chapter 14 Mathematical Reasoning Chapter 15 Statistics Chapter 16 Probability

What is meant by Case Study Question?

In the context of CBSE (Central Board of Secondary Education), a case study question is a type of question that requires students to analyze a given scenario or situation and apply their knowledge and skills to solve a problem or answer a question related to the case study.

Case study questions typically involve a real-world situation that requires students to identify the problem or issue, analyze the relevant information, and apply their understanding of the relevant concepts to propose a solution or answer a question. These questions may involve multiple steps and require students to think critically, apply their problem-solving skills, and communicate their reasoning effectively.

Importance of Solving Case Study Questions for Class 11 Maths

Case study questions are an important aspect of mathematics education at the Class 11 level. These questions require students to apply their knowledge and skills to real-world scenarios, helping them develop critical thinking, problem-solving, and analytical skills. Here are some reasons why case study questions are important in Class 11 maths education:

• Real-world application: Case study questions allow students to see how the concepts they are learning in mathematics can be applied in real-life situations. This helps students understand the relevance and importance of mathematics in their daily lives.
• Higher-order thinking: Case study questions require students to think critically, analyze data, and make connections between different concepts. This helps develop higher-order thinking skills, which are essential for success in both academics and real-life situations.
• Collaborative learning: Case study questions often require students to work in groups, which promotes collaborative learning and helps students develop communication and teamwork skills.
• Problem-solving skills: Case study questions require students to apply their knowledge and skills to solve complex problems. This helps develop problem-solving skills, which are essential in many careers and in everyday life.
• Exam preparation: Case study questions are included in exams and tests, so practicing them can help students prepare for these assessments.

Overall, case study questions are an important component of Class 11 mathematics education, as they help students develop critical thinking, problem-solving, and analytical skills, which are essential for success in both academics and real-life situations.

Feature of Case Study Questions on This Website

Here are some features of a Class 11 Maths Case Study Questions Booklet:

Many Case Study Questions: This website contains many case study questions, each with a unique scenario and problem statement.

Different types of problems: The booklet includes different types of problems, such as optimization problems, application problems, and interpretation problems, to test students’ understanding of various mathematical concepts and their ability to apply them to real-world situations.

Multiple-choice questions: Questions contains multiple-choice questions to assess students’ knowledge, understanding, and critical thinking skills.

Focus on problem-solving skills: The questions are designed to test students’ problem-solving skills, requiring them to identify the problem, select appropriate mathematical tools, and analyze and interpret the results.

Emphasis on practical applications: The case studies in the booklet focus on practical applications of mathematical concepts, allowing students to develop an understanding of how mathematics is used in real-life situations.

Comprehensive answer key: The booklet includes a comprehensive answer key that provides detailed explanations and step-by-step solutions for all the questions, helping students to understand the concepts and methods used to solve each problem.

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• Cube and Cube Roots Class 8 Assertion Reason Questions Maths Chapter 6

Last Updated on September 18, 2024 by XAM CONTENT

Hello students, we are providing case study questions for class 8 maths. Assertion Reason questions are the new question format that is introduced in CBSE board. The resources for assertion reason questions are very less. So, to help students we have created chapterwise assertion reason questions for class 8 maths. In this article, you will find assertion reason questions for CBSE Class 8 Maths Chapter 6 Cube and Cube Roots. It is a part of Assertion Reason Questions for CBSE Class 8 Maths Series.

Table of Contents

Assertion Reason Questions on Cube and Cube Roots

Questions :

Q. 1. Assertion (A): The one’s digit in the cube root of the cube number 1728 is 6. Reason (R): The cube root of a number is the factor that we multiply by itself three times to get that number (a) Both A and R are true and R is the correct explanation of A. (b) Both A and R are true but R is not the correct explanation of A. (c) A is true but R is false. (d) A is false but R is true.

Difficulty Level: Medium

Ans. Option (d) is correct Explanation: one’s digit in the cube root of cube number is 2 so, A is false but R is true.

Q. 2. Assertion (A): The smallest number by which the number 72 must be multiplied to obtain a perfect cube is 3. Reason (R): The perfect cube is the result of multiplying the same integer three times. (a) Both A and R are true and R is the correct explanation of A. (b) Both A and R are true but R is not the correct explanation of A. (c) A is true but R is false. (d) A is false but R is true.

Ans. Option (a) is correct Explanation: We can obtain cube root of 72 only when we multiply 72 by 3 because after multiplication we get 216 whose cube root is 6. And according to reason if we multiply 6 x 6 x 6 we get 216. Hence, both A and R are true.

• Comparing Quantities Class 8 Assertion Reason Questions Maths Chapter 7
• Algebraic Expressions and Identities Class 8 Assertion Reason Questions Maths Chapter 8
• Square and Square Roots Class 8 Assertion Reason Questions Maths Chapter 5
• Data Handling Class 8 Assertion Reason Questions Maths Chapter 4
• Understanding Quadrilaterals Class 8 Assertion Reason Questions Maths Chapter 3

Linear Equations in One Variable Class 8 Assertion Reason Questions Maths Chapter 2

Rational numbers class 8 assertion reason questions maths chapter 1, you may also like.

• Case Study Questions for CBSE Class 8 Maths

Topics from which assertion reason questions may be asked

• Adding Consecutive Odd Numbers
• Cubes and their Prime Factors
• Smallest multiple that is a perfect cube.
• Cube Root through Prime Factorisation method

Cube of an even natural numbers are even. Cube of an odd natural numbers are odd. The cubes of numbers ending with digits 0, 1, 4, 5, 6 and 9 end with same digits. Cube of three numbers, 0, 1 and -1 is equal to number itself.

Assertion reason questions from the above given topic may be asked.

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Frequently Asked Questions (FAQs) on Cube and Cube Roots Assertion Reason Questions Class 8

Q1: what are assertion reason questions.

A1: Assertion-reason questions consist of two statements: an assertion (A) and a reason (R). The task is to determine the correctness of both statements and the relationship between them. The options usually include: (i) Both A and R are true, and R is the correct explanation of A. (ii) Both A and R are true, but R is not the correct explanation of A. (iii) A is true, but R is false. (iv) A is false, but R is true. or A is false, and R is also false.

Q2: Why are assertion reason questions important in Maths?

A2: Students need to evaluate the logical relationship between the assertion and the reason. This practice strengthens their logical reasoning skills, which are essential in mathematics and other areas of study.

Q3: How can practicing assertion reason questions help students?

A3: Practicing assertion-reason questions can help students in several ways: Improved Conceptual Understanding:  It helps students to better understand the concepts by linking assertions with their reasons. Enhanced Analytical Skills:  It enhances analytical skills as students need to critically analyze the statements and their relationships. Better Exam Preparation:  These questions are asked in exams and practicing them can improve your performance.

Q4: What strategies should students use to answer assertion reason questions effectively?

A4: Students can use the following strategies: Understand Each Statement Separately:  Determine if each statement is true or false independently. Analyze the Relationship:  If both statements are true, check if the reason correctly explains the assertion.

Q5: What are common mistakes to avoid when answering Assertion Reason questions?

A5: Common mistakes include: Not reading the statements carefully and missing key details. Assuming the Reason explains the Assertion without checking the logical connection. Confusing the order or relationship between the statements. Overthinking and adding information not provided in the question.

Q6: What resources can help me practice Assertion Reason Questions for Class 8 Maths?

A6: Use study guides specifically designed for Assertion-Reason questions. Online educational platforms and reference books for Class 8 Maths also offer practice questions and explanations. xamcontent.com also provides assertion reason questions for cbse class 8 maths.

Q7: What are cube roots?

A7: A cube root is the value that, when multiplied by itself three times, gives the original number.

Q8: How can you determine if a number is a perfect cube?

A8: A number is a perfect cube if its cube root is an integer. For example, 64 is a perfect cube because cube root of 64 is 4, which is an integer. To check if a number is a perfect cube, find its prime factorization and see if the exponent of each prime factor is a multiple of 3.

Q9: What is the relationship between cube and cube root?

A9: The cube of a number is the result of raising that number to the power of three, while the cube root is the reverse process, finding the number that was cubed.

Q10: Are there any online resources or tools available for practicing “Cube and Cube Roots “ assertion reason questions?

A10: A9: We provide assertion reason questions for CBSE Class 8 Maths on our  website . Students can visit the website and practice sufficient assertion reason questions and prepare for their exams. If you need more case study questions, then you can visit  Physics Gurukul  website. they are having a large collection of case study questions for all classes.

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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].

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