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.

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MCQs for Chapter 3 Trigonometric Functions Class 11 with Answers

1. If sin θ and cos θ are the roots of ax2 – bx + c = 0, then the relation between a, b and c will be

(a) a2 + b2 + 2ac = 0

(b) a2 – b2 + 2ac = 0

(c) a2 + c2 + 2ab = 0

(d) a2 – b2 – 2ac = 0

Correct option: (b) a2 – b2 + 2ac = 0

Solution:

Given that sin θ and cos θ are the roots of the equation ax2 – bx + c = 0, so sin θ + cos θ = b/a

and sin θ cos θ = c/a

Consider,

(sinθ + cos θ)2 = sin2θ + cos2θ + 2 sin θ cos θ,

(b/a)2 = 1 + 2(c/a) {using the identity sin2A + cos2A = 1}

b2/a2 = 1 + (2c/a)

b2 = a2 + 2ac

a2 – b2 + 2ac = 0

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

(a) π/6

(b) π

(c) 0

(d) π/4

Correct option: (d) π/4

Solution:

Given,

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

We know that,

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

= [(1/2) + (1/3)]/ [1 – (1/2)(1/3)]

= [(3 + 2)/6]/ [(6 – 1)/6]

= 5/5

= 1

= tan π/4

Therefore, A + B = π/4

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

(a) 1/√2

(b) 0

(c) 1

(d) –1

Correct option: (b) 0

Solution:

cos 1° cos 2° cos 3° … cos 179°

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

(a) 1

(b) 0

(c) 1/2

(d) 2

Correct option: (b) 0

Solution:

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°

= 0

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

(a) 2 cosθ

(b) 2 sinθ

(c) 1

(d) 0

Correct option: (d) 0

Solution:

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)

= 0

Alternative method:

sin (45° + θ) – cos (45° – θ)

Using the formulas

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

= 0

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

(a) 0

(b) 1

(c) 1/2

(d) Not defined

Correct option: (b) 1

Solution:

tan 1° tan 2° tan 3° … tan 89°

= [tan 1° tan 2° … tan 44°] tan 45°[tan (90° – 44°) tan (90° – 43°)… tan (90° – 1°)]

= [tan 1° tan 2° … tan 44°] [cot 44° cot 43°……. cot 1°] × [tan 45°]

= [(tan 1° × cot 1°) (tan 2° × cot 2°)…..(tan 44° × cot 44°)] × [tan 45°]

We know that,

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

(a) 1

(b) 2

(c) – 2

(d) Not defined

Correct option: (b) 2

Solution:

Given,

α + β = π/4

Taking “tan” on both sides,

tan(α + β) = tan π/4

We know that,

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

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

= 1 + 1 [From (i)]

= 2

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

(a) -53/10

(b) 23/10

(c) 37/10

(d) 7/10

Correct option: (b) 23/10

Solution:

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 sec2A = 1 + tan2A,

sec2A = 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 sin2A + cos2A = 1,

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

Now,

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

= 23/10

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

Solution:

Given,

cos θ = x + (1/x)

cos θ = (x2 + 1)/x

⇒ x2 + 1 = x cos θ

⇒ x2 – x cos θ + 1 = 0

We know that for any real root of the equation ax2 + bx + c = 0, b2 – 4ac ≥ 0.

⇒ (-cos θ)2 – 4 ≥ 0

⇒ cos2θ – 4 ≥ 0

⇒ cos2θ ≥ 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

(a) 0

(b) 1

(c) 2

(d) 3

Correct option: (c) 2

Solution:

Given,

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 cos2x

Using the identity sin2A + cos2A = 1,

sin x + 1 = 2(1 – sin2x)

sin x + 1 = 2 – 2 sin2x

⇒ 2 sin2x + sin x − 1 = 0

⇒ (2 sin x − 1)(sin x + 1)=0

⇒ sin x = −1, 1/2

⇒ x = π/6,5π/6,3π/2 ∈ [0, 2π]

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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