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 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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