# RBSE Solutions for Class 10 Maths Chapter 6 Trigonometric Ratios Miscellaneous Exercise

RBSE Solutions for Class 10 Maths Chapter 6 Trigonometric Ratios Miscellaneous Exercise is part of RBSE Solutions for Class 10 Maths. Here we have given Rajasthan Board RBSE Class 10 Maths Chapter 6 Trigonometric Ratios Miscellaneous Exercise.

## Rajasthan Board RBSE Class 10 Maths Chapter 6 Trigonometric Ratios Miscellaneous Exercise

Multiple Choice Questions
RBSE Class 10 Maths Chapter 6 Miscellaneous Question 1.
The value (RBSESolutions.com) of tan2 60° is :
(A) 3
(B) $$\frac { 1 }{ 3 }$$
(C) 1
(D) ∞
Solution :
tan2 60° = (√3)2 = 3
(A) is correct.

RBSE Solutions For Class 10 Maths Chapter 6 Miscellaneous Question 2.
The value (RBSESolutions.com) of 2 sin2 60° cos 60° is :
(A) $$\frac { 4 }{ 3 }$$
(B) $$\frac { 5 }{ 2 }$$
(C) $$\frac { 3 }{ 4 }$$
(D) $$\frac { 1 }{ 3 }$$
Solution :
2 sin2 60° cos 60° Hence (C) is correct.

RBSE Solutions For Class 10 Maths Chapter 6 Question 3.
If cosec θ = $$\frac { 2 }{ \sqrt { 3 } }$$ then (RBSESolutions.com) value of θ is : Solution :
cosec θ = $$\frac { 2 }{ \sqrt { 3 } }$$
cosec 60° = cosec $$\frac { \pi }{ 3 }$$
θ = $$\frac { \pi }{ 3 }$$
Hence, (B) is correct..

Class 10 Maths RBSE Solution Chapter 6 Question 4.
The value of cos2 = 45° is : Solution :
cos 45° = $${ \left( \frac { 1 }{ \sqrt { 2 } } \right) }^{ 2 }$$ = $$\frac { 1 }{ 2 }$$
Hence, (C) is correct.

RBSE Maths Solution Class 10 Chapter 6 Question 5.
If θ = 45° then (RBSESolutions.com) value of $$\frac { 1-cos2\theta }{ sin2\theta }$$ is :
(A) 0
(B) 1
(C) 2
(D) ∞
Solution : Hence, (B) is (RBSESolutions.com) correct.

Miscellaneous Exercise Class 10 Question 6.
cos 60° = 2 cos2 30° – 1
Solution :
R.H.S. L.H.S. = cos 60° = $$\frac { 1 }{ 2 }$$
∴ L.H.S. = R.H.S.

RBSE Solution Class 10 Maths Chapter 6 Question 7. Solution :
R.H.S. L.H.S.

Maths Class 10th RBSE Solution Question 8. Solution :
R.H.S. L.H.S.

RBSE Solution Class 10th Maths Question 9.
(sin 45° + cos 45°)2 = 2
Solution :
L.H.S.
= (sin 45° + cos 45°)2 = 2 = R.H.S.

RBSE Solutions Class 10 Maths Question 10.
tan 30° sin 45° (RBSESolutions.com) sin 60° sin 90° = √2
Solution :
L.H.S. = 4 tan 30° sin 45° sin 60° sin 90° = R.H.S.

Class 10th Maths Ncert Solutions Chapter 6 Question 11.
Evaluate : sin2 60° cot2 60°.
solution :
sin2 60° cot2 60° Class 10 Maths RBSE Solution Question 12.
Evaluate :
4 cos3 30° – 3 cos 30°.
Solution :
4 cos3 30° – 3 cos 30° RBSE Solution Class 10 Maths Question 13.
If cot θ = $$\frac { 1 }{ \sqrt { 3 } }$$ then (RBSESolutions.com) prove that Solution :
Given, cot θ = $$\frac { 1 }{ \sqrt { 3 } }$$
cot θ = cot 60°
∴ θ = 60° RBSE Solutions For Class 10th Maths Question 14.
Prove (RBSESolutions.com) that 3(tan2 30° + cot2 30°) – 8(sin2 45° + cos2 30°) = 0
Solution :
3(tan2 30° + cot2 30°) – 8(sin2 45° + cos2 30°) = 10 – 10 = 0
Thus, 3(tan2 30° + cot2 30°) – 8(sin2 45° + cos2 30°) = 0

Chapter 6 Maths Class 10 Question 15.
4(sin4 30° + cos 60°) – 3(cos2 45° – sin2 90°) = $$\frac { 15 }{ 4 }$$
Solution :
4(sin4 30° + cos 60°) – 3(cos2 45° – sin2 90°) Thus, 4(sin4 30° + cos 60°) – 3(cos2 45° – sin2 90°) = $$\frac { 15 }{ 4 }$$

Ch 6 Maths Class 10 Question 16. Solution : Chapter 6 Class 10 Maths Question 17.
2(cos2 45° + tan2 60°) – 6(sin2 45° – tan2 30°) = 6
Solution :
2(cos2 45° + tan2 60°) – 6(sin2 45° – tan2 30°) = 2 × $$\frac { 7 }{ 2 }$$ – 6 × $$\frac { 1 }{ 6 }$$ = 7 – 1 = 6
Thus, 2(cos2 45° + tan2 60°) – 6(sin2 45° – tan2 30°) = 6

We hope the given RBSE Solutions for Class 10 Maths Chapter 6 Trigonometric Ratios Miscellaneous Exercise will help you. If you have any query regarding Rajasthan Board RBSE Class 10 Maths Chapter 6 Trigonometric Ratios Miscellaneous Exercise, drop a comment below and we will get back to you at the earliest.