Trigonometric Functions - Class 11 Maths MCQ

This set of Class 11 Maths Chapter 3 Multiple Choice Questions & Answers (MCQs) focuses on “Trigonometric Functions – 2”.

1. cosec(-30°) =___________
a) -2
b) 2
c) 2/√3
d) -2/√3
View Answer

Answer: a
Explanation: We know, cosec(-x) = -cosec(x)
So, cosec(-30°) = -cosec(30°) = -2.

2. sec(-45°) =_____________
a) 1
b) -1
c) √2
d) -√2
View Answer

Answer: c
Explanation: We know, sec(-x) = sec x
So, sec(-45°)=sec 45°=1/(cos 45°)=√2.

3. cot x is not defined for_______
a) 0
b) nπ/2
c) (2n+1) π/2
d) nπ
View Answer

Answer: d
Explanation: We know, cot x is not defined when sin x = 0.
sin x = 0 whenever x is 0, π, 2π, 3π, …. i.e. all integral multiples of π
so, x=nπ.

4. If sin x=-4/5 and x lies in 3^rd quadrant, then find sec x.
a) 5/3
b) 3/5
c) -3/5
d) -5/3
View Answer

Answer: d
Explanation: sin x=-4/5
We know, sin²x + cos²x=1
cos²x = 1-(-4/5)² = 1-16/25=9/25
cos x=±3/5
cos x is negative in 3^rd quadrant so, cos x=-3/5.
sec x = 1/cos x = 1/(-3/5) = – 5/3.

5. If cosec x = -5/12 and x lies in 2^nd quadrant, then find cos x.
a) 12/13
b) 5/13
c) -13/5
d) 12/13
View Answer

Answer: c
Explanation: cosec x = 5/12
sin x = 1/cosec x = 12/5
We know, sin²x+cos²x=1
cos² x = 1-(-12/5)² = 1+144/25 = 169/25
cos x = ±13/5
cos x is negative in 2^nd quadrant so, cos x=-13/5.

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6. If sec x = 13/5 and x lies in 4^th quadrant, then find cot x.
a) 5/12
b) -5/12
c) 5/13
d) -5/13
View Answer

Answer: b
Explanation: sec x = 13/5.
We know, sec²x – tan²x=1
tan²x = (13/5)²-1 = (169/25) – 1 = 144/25
tan x = ±12/5
tan x is negative in 4^th quadrant so, tan x=-12/5
cot x = 1/tan x = 1/(-12/5) = – 5/12.

7. If tan x = -5/12 and x lies in 2^nd quadrant, then find cosec x.
a) 12/5
b) 13/5
c) -13/5
d) -12/5
View Answer

Answer: b
Explanation: tan x = -5/12
cot x = 1/tan x = 1/ (-5/12) = -12/5
We know, cosec²x – cot²x = 1
cosec²x =1+(-12/5)² = 1+144/25 = 169/25
cosec x = ± 13/5
cosec x is positive in 2^nd quadrant so, cosec x = 13/5.

8. sin (15π/6) =_____________
a) 1
b) -1
c) 0
d) 1/2
View Answer

Answer: a
Explanation: sin(15π/6) = sin (2π + 3π/6) = sin (3π/6) {sin(2nπ+x)=sin x}
= sin (π/2) = 1.

9. cos (17π/3) =______________
a) 1/2
b) -1/2
c) √3/2
d) -√3/2
View Answer

Answer: a
Explanation: cos (17π/3) = cos (2π*3 – π/3)
{cos (2nπ-x)=cos x}
= cos(π/3) = 1/2.

10. tan (19π/6) =____________________
a) √3
b) -1/√3
c) – √3
d) 1/√3
View Answer

Answer: d
Explanation: tan (19π/6) = tan (2π + 7π/6) = tan 7π/6 {tan( 2nπ+x)=tan x}
= tan (π+π/6) = tan π/6 = 1/√3. {tan π+x = tan x}

11. cos ( -1500°) =______________
a) 1/2
b) -1/2
c) √3/2
d) -√3/2
View Answer

Answer: a
Explanation: cos(-1500°) = cos(1500°) {cos(-x) = cos x}
= cos (4*360° + 60°) = cos 60° = 1/2. {cos (2nπ+x) = cos x}

12. sin 1710° =__________________
a) 1
b) -1
c) 0
d) 1/2
View Answer

Answer: b
Explanation: sin 1710° = sin (360°*5 – 90°) {sin (2nπ-x)= – sin x}
=-sin 90° = -1.

13. tan 1560°=_________________
a) -√3
b) √3
c) 1/√3
d) -1/√3
View Answer

Answer: a
Explanation: tan 1560° = tan (360°*4 + 120°) = tan 120° {tan (2nπ+x) = tan x}
= tan (180°-60°) = -tan 60° = – √3. {tan π-x = – tan x}

Sanfoundry Global Education & Learning Series – Mathematics – Class 11.

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