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
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.

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.

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..

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.

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.

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.

Question 7.

Solution :
R.H.S.

L.H.S.

Question 8.

Solution :
R.H.S.

L.H.S.

Question 9.
(sin 45° + cos 45°)² = 2
Solution :
L.H.S.
= (sin 45° + cos 45°)²

= 2 = R.H.S.

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.

Question 11.
Evaluate : sin² 60° cot² 60°.
solution :
sin² 60° cot² 60°

Question 12.
Evaluate :
4 cos³ 30° – 3 cos 30°.
Solution :
4 cos³ 30° – 3 cos 30°

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°

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

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°) = \(\frac { 15 }{ 4 }\)

Question 16.

Solution :

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.