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

May 3, 2019 by Fazal Leave a Comment

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°
RBSE Solutions for Class 10 Maths Chapter 6 Trigonometric Ratios Miscellaneous Exercise 1
Hence (C) is correct.

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

RBSE Solutions

Question 4.
The value of cos2 = 45° is :
RBSE Solutions for Class 10 Maths Chapter 6 Trigonometric Ratios Miscellaneous Exercise 3
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 :
RBSE Solutions for Class 10 Maths Chapter 6 Trigonometric Ratios Miscellaneous Exercise 4
Hence, (B) is (RBSESolutions.com) correct.

Question 6.
cos 60° = 2 cos2 30° – 1
Solution :
R.H.S.
RBSE Solutions for Class 10 Maths Chapter 6 Trigonometric Ratios Miscellaneous Exercise 5
L.H.S. = cos 60° = \(\frac { 1 }{ 2 }\)
∴ L.H.S. = R.H.S.

Question 7.
RBSE Solutions for Class 10 Maths Chapter 6 Trigonometric Ratios Miscellaneous Exercise 6
Solution :
R.H.S.
RBSE Solutions for Class 10 Maths Chapter 6 Trigonometric Ratios Miscellaneous Exercise 7
L.H.S.

Question 8.
RBSE Solutions for Class 10 Maths Chapter 6 Trigonometric Ratios Miscellaneous Exercise 8
Solution :
R.H.S.
RBSE Solutions for Class 10 Maths Chapter 6 Trigonometric Ratios Miscellaneous Exercise 9
L.H.S.

Question 9.
(sin 45° + cos 45°)2 = 2
Solution :
L.H.S.
= (sin 45° + cos 45°)2
RBSE Solutions for Class 10 Maths Chapter 6 Trigonometric Ratios Miscellaneous Exercise 10
= 2 = R.H.S.

RBSE Solutions

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°
RBSE Solutions for Class 10 Maths Chapter 6 Trigonometric Ratios Miscellaneous Exercise 11
= R.H.S.

Question 11.
Evaluate : sin2 60° cot2 60°.
solution :
sin2 60° cot2 60°
RBSE Solutions for Class 10 Maths Chapter 6 Trigonometric Ratios Miscellaneous Exercise 12

Question 12.
Evaluate :
4 cos3 30° – 3 cos 30°.
Solution :
4 cos3 30° – 3 cos 30°
RBSE Solutions for Class 10 Maths Chapter 6 Trigonometric Ratios Miscellaneous Exercise 13

Question 13.
If cot θ = \(\frac { 1 }{ \sqrt { 3 } }\) then (RBSESolutions.com) prove that
RBSE Solutions for Class 10 Maths Chapter 6 Trigonometric Ratios Miscellaneous Exercise 14Solution :
Given, cot θ = \(\frac { 1 }{ \sqrt { 3 } }\)
cot θ = cot 60°
∴ θ = 60°
RBSE Solutions for Class 10 Maths Chapter 6 Trigonometric Ratios Miscellaneous Exercise 15

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°)
RBSE Solutions for Class 10 Maths Chapter 6 Trigonometric Ratios Miscellaneous Exercise 16
= 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°)
RBSE Solutions for Class 10 Maths Chapter 6 Trigonometric Ratios Miscellaneous Exercise 17
Thus, 4(sin4 30° + cos 60°) – 3(cos2 45° – sin2 90°) = \(\frac { 15 }{ 4 }\)

RBSE Solutions

Question 16.
RBSE Solutions for Class 10 Maths Chapter 6 Trigonometric Ratios Miscellaneous Exercise 1
Solution :
RBSE Solutions for Class 10 Maths Chapter 6 Trigonometric Ratios Miscellaneous Exercise 19

Question 17.
2(cos2 45° + tan2 60°) – 6(sin2 45° – tan2 30°) = 6
Solution :
2(cos2 45° + tan2 60°) – 6(sin2 45° – tan2 30°)
RBSE Solutions for Class 10 Maths Chapter 6 Trigonometric Ratios Miscellaneous Exercise 20
= 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.

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