Practice Questions with Answers - Multiple Choice Questions

MCQ Questions for Class 11 Maths Chapter 3 Trigonometric Functions

MCQ Questions for Class 11 Maths Chapter 3 Trigonometric Functions

Students are advised to solve the Trigonometric Functions Multiple Choice Questions of Class 11 Maths to know different concepts. Practicing the MCQ Questions on Trigonometric Functions Class 11 with answers will boost your confidence thereby helping you score well in the exam.

Explore numerous MCQ Questions of Trigonometric Functions Class 11 with answers provided with detailed solutions by looking below.

Question 1.
The value of sin 15 + cos 15 is
(a) 1
(b) 1/2
(c) √3/2
(d) √3

Given, sin 15 + cos 15
= sin 15 + cos(90 – 15)
= sin 15 + sin 15
= 2 × sin 45 × cos 30
= 2 × (1/√2) × (√3/2)
= √3/2

Question 2.
The value of tan A/2 – cot A/2 + 2cot A is
(a) 0
(b) 1
(c) -1
(d) None of these

Given, tan A/2 – cot A/2 + 2cot A
= {sin(A/2)/cos(A/2)} – {cos(A/2)/sin(A/2)} + 2cotA
= {sin² (A/2) – cos² (A/2)}/{cos(A/2) × sin(A/2)} + 2cotA
= -{cos² (A/2) – sin² (A/2)}/{cos(A/2) × sin(A/2)} + 2cotA
= -{cosA}/{cos(A/2) × sin(A/2)} + 2cotA (since cos² A – sin² A = cos²A )
= -{cos(2A/2)}/{cos(A/2) × sin(A/2)} + 2cotA
= -{2 × cosA}/{2 × cos(A/2) × sin(A/2)} + 2cotA
= -{2cosA}/{sin(2A/2)} + 2cotA
= {-(2cosA)/(sinA)} + 2cotA (since sin2A = 2 × sinA × cosA)
= -2cotA + 2cotA
= 0

Question 3.
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

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 × {-(sin² x)/4 + (3 × cos² x)/4}
= sin x × {-sin² x + 3 × cos² x}
= sin x × {-sin² x + 3 × (1 – sin² x)}
= sin x × {-sin² x + 3 – 3 × sin² x}
= sin x × {3 – 4 × sin² x}
= 3 × sin x – 4sin³ x
= sin 3x
So, 4 × sin x × sin(x + π/3) × sin(x + 2π/3) = sin 3x

Question 4.
If tan x = (cos 9 + sin 9)/(cos 9 – sin 9), then x =
(a) 45
(b) 54
(c) 36
(d) None of these

Given, tan x = (cos 9 + sin 9)/(cos 9 – sin 9)
⇒ tan x = {cos 9(1 + sin 9/cos 9)}/{cos 9(1 – sin 9/cos 9)}
⇒ tan x = (1 + tan 9)}/(1 – tan 9)
⇒ tan x = (tan 45 + tan 9)}/(1 – tan 45 × tan 9) {since tan 45 = 1}
⇒ tan x = tan(45 + 9) {Apply tan(A + B) formula}
⇒ tan x = tan(54)
⇒ x = 54

Question 5.
In a triangle ABC, sin A – cos B = cos C, then angle B is
(a) π/2
(b) π/3
(c) π/4
(d) π/6

Given, sin A – cos B = sin C
⇒ sin A = cos B + sin C
⇒ 2 × sin (A/2) × cos (A/2) = 2 × cos {(B + C)/2} × cos {(B – C)/2}
⇒ 2 × sin (A/2) × cos (A/2) = 2 × cos {π/2 – A/2} × cos {(B – C)/2}
⇒ 2 × sin (A/2) × cos (A/2) = 2 × sin (A/2) × cos {(B – C)/2}
⇒ cos (A/2) = cos {(B – C)/2}
⇒ A/2 = (B – C)/2
⇒ A = B – C
⇒ B = A + C
⇒ B = π – B {Since A + B + C = π}
⇒ 2B = π
⇒ B = π/2

Question 6.
The value of cos 420° is
(a) 0
(b) 1
(c) 1/2
(d) √3/2

cos 420° = cos(360° + 60° ) = cos 60° = 1/2

Question 7.
If in a triangle ABC, tan A + tan B + tan C = 6 then the value of cot A × cot B × cot C is
(a) 1/2
(b) 1/3
(c) 1/4
(d) 1/6

Given tanA + tanB + tanC = 6
Now tan(A + B + C) = {(tanA + tanB + tanC) – tanA × tanB × tanC}/{1 – (tanA × tanB + tanB × tanC + tanA × tanC)}
We know that,
A + B + C = π
⇒ tan(A + B + C) = tan π
⇒ tan(A + B + C) = 0
Now
0 = {(tanA + tanB + tanC) – tanA × tanB × tanC}/{1 – (tanA × tanB + tanB × tanC + tanA × tanC)}
⇒ tanA + tanB + tanC – tanA × tanB × tanC = 0
⇒ tanA + tanB + tanC = tanA × tanB × tanC
⇒ tanA × tanB × tanC = 6
⇒ (1/cotA) × (1/cotB) × (1/cotC) = 6
⇒ 1/(cot A × cot B × cot C) = 6
⇒ cot A × cot B × cot C = 1/6

Question 8.
If a × cos x + b × cos x = c, then the value of (a × sin x – b × cos x)² is
(a) a² + b² + c²
(b) a² – b² – c²
(c) a² – b² + c²
(d) a² + b² – c²

Answer: (d) a² + b² – c²
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 9.
When the length of the shadow of a pole is equal to the height of the pole, then the elevation of source of light is
(a) 30°
(b) 60°
(c) 75°
(d) 45°

Let AB is the length of the pole and BC is the shadow of the pole.
Given AB = BC
Now from triangle ABC,
tan θ = AB/BC
⇒ tan θ = 1
⇒ θ = 45°
So, the elevation of source of light is 45°

Question 10.
In any triangle ABC, if cos A/a = cos B/b = cos C/c and the side a = 2, then the area of the triangle is
(a) √3
(b) √3/4
(c) √3/2
(d) 1/√3