# RBSE Solutions for Class 11 Maths Chapter 3 Trigonometric Functions Miscellaneous Exercise

## Rajasthan Board RBSE Class 11 Maths Chapter 3 Trigonometric Functions Miscellaneous Exercise

Question 1.
A right angle is:
(B) equal to 90 degree
(C) equal to 18°
Solution :
(B) equal to 90 degree

Question 2.
Which trigonometric function is positive in third quadrant:
(A) sin θ
(B) tan θ
(C) cos θ
(D) sec θ
Solution:
(B) tan θ

Question 3.
cosec (- θ) is equal to:
(A) sin θ
(B) tan θ
(C) cos θ
(D) -cosec θ
Solution:
(D) -cosec θ

Question 4.
tan (90° – θ) is equal to:
(A) -tan θ
(B) cot θ
(C) tan θ
(D) -cot θ
Solution :
(B) cot θ

Question 5.
If cos θ = $$\frac { 1 }{ 2 }$$ then value of θ will be:
(A) $$\frac { 2\pi }{ 3 }$$
(B) $$\frac { \pi }{ 3 }$$
(C) –$$\frac { 2\pi }{ 3 }$$
(D) $$\frac { 3\pi }{ 4 }$$
Solution:
(A) $$\frac { 2\pi }{ 3 }$$

Question 6.
If n is an even integer, then the value of sin (2nπ ± θ) will be:
(A) ± cos θ
(B) ± tan θ
(C) ± sin θ
(D) ± cot θ
Solution:
(C) ± sin θ

Question 7.
The value of cot 15° will be:
(A) 2 + √3
(B) – 2 + √3
(C) 2 – √3
(D) – 2 – √3
Solution:
(A) 2 + √3

Question 8.
The value of cos 15° will be:
(A) $$\frac { \sqrt { 3 } +1 }{ 2\sqrt { 2 } }$$
(B) $$\frac { \sqrt { 3 } -1 }{ 2 }$$
(C) $$\frac { \sqrt { 3 } -1 }{ 2\sqrt { 2 } }$$
(D) $$\frac { \sqrt { 3 } +1 }{ 2 }$$
Solution:
(A) $$\frac { \sqrt { 3 } +1 }{ 2\sqrt { 2 } }$$

Question 9.
The value of 2 sin $$\frac { 5\pi }{ 12 }$$ cos $$\frac { \pi }{ 12 }$$ will be:
(A) 1
(B) $$\frac { \sqrt { 3 } }{ 2 }$$
(C) $$\frac { \sqrt { 3 } }{ 2 }$$ -1
(D) $$\frac { \sqrt { 3 } }{ 2 }$$ +1
Solution:
(D) $$\frac { \sqrt { 3 } }{ 2 }$$ +1
2 sin $$\frac { 5\pi }{ 12 }$$ cos $$\frac { \pi }{ 12 }$$
= 2 sin 75° cos 15°
= 2 sin (90° – 15°) cos 15°
= 2 cos 15° cos 15°
= 2 cos2 15°
= cos2 × 15° + 1
= cos 30° + 1
= $$\frac { \sqrt { 3 } }{ 2 }$$ +1

Question 10.
The value of cos $$\frac { 5\pi }{ 12 }$$ sin $$\frac { \pi }{ 12 }$$ will be:
(A) $$\frac { 1 } { 2\sqrt { 2 } }$$
(B) 0
(C) $$\frac -{ 1 } { 2\sqrt { 2 } }$$
(D) $$\frac { 1 } { \sqrt { 2 } }$$
Solution:
(D) $$\frac { 1 } { \sqrt { 2 } }$$
cos $$\frac { \pi }{ 12 }$$ – sin $$\frac { \pi }{ 12 }$$ = cos 15°- sin 15°
= cos (45° – 30°) – sin (45° – 30°)
= (cos 45° cos 30° + sin 45° sin 30°)
= (sin 45° cos 30° – cos 45° sin 30°)

Question 11.
If sin A = $$\frac { 3 }{ 5 }$$ than value sin 2A will be:
(A) $$\frac { 3 }{ 5 }$$
(B) $$\frac { 5 }{ 25 }$$
(C) $$\frac { 24 }{ 25 }$$
(D) $$\frac { 4 }{ 5 }$$
solution:

Hence,Option (C) is correct.

Question 12.
If sin A = $$\frac { 3 }{ 4 }$$ than value of sin 3A will be:
(A) $$\frac { 9 }{ 16 }$$
(B) – $$\frac { 9 }{ 16 }$$
(C) $$\frac { 9 }{ 32 }$$
(D) $$\frac { 7 }{ 16 }$$
Solution:
sin 3A
3 sin A – 4 sin3 A

Hence,Option (A) is correct.

Question 13.
If tan A = $$\frac { 1 }{ 5 }$$ then value of tan 3A will be:
(A) $$\frac { 47 }{ 25 }$$
(B) $$\frac { 37 }{ 55 }$$
(C) $$\frac { 37 }{ 11 }$$
(D) $$\frac { 47 }{ 55 }$$
Solution:

Hence,Option (B) is correct.

Question 14.
If A + B = $$\frac { \pi }{ 4}$$ then value (1 + tan A) (1 + tan B) will be:
(A) 3
(B) 2
(C) 4
(D) 1
Solution:
(1 + tan A) (1 + tan B)
= 1 + tan A + tan B + tan A tan B ….. (i)

From equation (1) and (2)
(1 + tan A) (1 + tan B) = 2.

Question 15.
General value of θ in equation sec2 θ = 2 will be:

Solution:

Hence,Option (A) is correct.

Question 16.
Prove that:
(i) cos θ + sin (27θ° + θ) – sin (27θ° – θ)+ cos (18θ° + θ) = θ

Solution:
(i) sin (27θ° + θ) → IV quardant, -ive = -cosθ … (i)
sin (27θ° – θ) → III quardant, -ive = – cos θ …. (ii)
cos (18θ° + θ) → III quardrant, -ive = – cos θ … (iii)
L.H.S. = cos θ + sin (27θ° + θ) – sin (27θ° – θ) + cos (18θ° + 9)
= cos θ – cos θ + cos θ – cos θ [from (i), (ii) and (iii)]
= θ
= R.H.S.
Hence Proved.

= sec (27θ° – θ) sec (θ – 45θ°)+ tan (45θ° + θ) tan (θ – 27θ°)
= – cosec θ sec (θ – 45θ°)+ tan (36θ° + 9θ° + θ) tan (θ – 27θ°)
= – cosec θ sec (45θ° – θ) – tan (9θ° + θ) tan (27θ° – θ)
= – cosec θ sec (36θ° + 9θ° – θ) – (- cot θ) cot θ
= – cosec θ sec (9θ° – θ) + cot2 θ
= – cosec θ cosec θ + cot2 θ
= – cosec2 θ + cot2 θ
= – 1 (∵ 1 + cot2 θ = cosec2 θ cot2 θ – cosec2 θ = – 1)
R.H.S.
Hence Proved.

Question 17.
Find the value of sin $$\left\{ n\pi +{ \left( -1 \right) }^{ n }\frac { \pi }{ 4 } \right\}$$, where n is an integer.
Solution:

Question 18.
If sin A + sin 5 = a and cos A + cos 5 = b, then prove that:
(i) sin (A + B) = $$\frac { \left( 2ab \right) }{ \left( { a }^{ 2 }+{ b }^{ 2 } \right) }$$
(ii) cos (A + B) = $$\frac { \left( { b }^{ 2 }-{ a }^{ 2 } \right) }{ \left( { a }^{ 2 }+{ b }^{ 2 } \right) }$$
Solution:

Question 19.
If A + B + C – 180°, then prove that:
(i) cos 2A + cos 2B – cos 2C = 1 – 4 sin A sin 5 cos C
(ii) sin A – sin B + sin C = 4 sin $$\frac { A }{ 2 }$$ cos $$\frac { B }{ 2 }$$ sin $$\frac { C }{ 2 }$$
Solution:
(i) L.H.S.
= cos 2A + cos 2B – cos 2C
= (cos 2A + cos 2B) – cos 2C
= 2 cos (A + B) cos (A – B) – cos 2C
= 2 cos (180° – C) cos (A – B) – (2 cos2 C – 1)
= -2 cos C cos (A – B) -2 cos2 C + 1
= 1 – 2 cos C cos (A – B) -2 cos2 C
= 1 – 2 cos C [cos (A – B) + cos C]

Question 20.
If A + B + C = 2π, then prove that cos2 B + cos2 C – sin2 A = 2 cos A cos B cos C.
Solution:
We know that:

= [cos2B + cos (2π + B) cos (A – C)]
= cos2B + cos B cos (A – C)
= cos B (cos B + cos (A – C)
= cos B (cos (2π – (A – C) + cos (A – C) – cos B [cos (A – B) + cos (A + C)]
= cos B [(cos (A + B) + cos (A – C)]
= cos B [2 cos A cos C]
= 2 cos A cos B cos C
Hence Proved

Question 21.
Find the following equation 2 tan θ – cot θ +1 = 0.
Solution:

When tan θ + 1 = 0
then tan θ = – 1 = tan 135° or tan $$\frac { 3\pi }{ 4 }$$
⇒ Hence, principal value of θ = $$\frac { 3\pi }{ 4 }$$
And general value of θ = nπ + $$\frac { 3\pi }{ 4 }$$ ….. (i)
when 2 tan θ – 1 =0
⇒ tan θ = $$\frac { 1 }{ 2 }$$
Hence,principal value of θ = tan-1$$\frac { 1 }{ 2 }$$
and general value of θ = nπ + tan-1$$\frac { 1 }{ 2 }$$ ….. (ii)
From (i) and (ii), solution of given equation is
nπ + tan-1$$\frac { 1 }{ 2 }$$ and nπ + $$\frac { 3\pi }{ 4}$$ , where n ∈ Z