Chapter 6 Trigonometric Ratios Miscellaneous Exercise

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

Class 10 Maths Chapter 6 Trigonometric Ratios Miscellaneous Exercise Solution is provided in this post. Here we have provide the solutions of RBSE Boards Books according to chapter wise.

Chapter 6 Trigonometric Ratios Miscellaneous Exercise

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

ultiple Choice Questions
Question 1.
The value of tan2 60° is :
(A) 3
(B) 
(C) 1
(D) ∞
Solution :
tan2 60° = (√3)2 = 3
(A) is correct.

Question 2.
The value of 2 sin2 60° cos 60° is :
(A) 
(B) 
(C) 
(D) 
Solution :
2 sin2 60° cos 60°

Hence (C) is correct.

Chapter 6 Trigonometric Ratios Miscellaneous Exercise

Question 3.
If cosec θ =  then value of θ is :

Solution :
cosec θ = 
cosec 60° = cosec 
θ = 
Hence, (B) is correct.

Question 4.
The value of cos2 = 45° is :

Solution :
cos 45° =  = 
Hence, (C) is correct.

Question 5.
If θ = 45° then value of  is :
(A) 0
(B) 1
(C) 2
(D) ∞
Solution :

Hence, (B) is correct.

Question 6.
cos 60° = 2 cos2 30° – 1
Solution :
R.H.S.

L.H.S. = cos 60° = 
∴ L.H.S. = R.H.S.

Chapter 6 Trigonometric Ratios Miscellaneous Exercise

Question 7.

Solution :
R.H.S.

L.H.S.

Question 8.

Solution :
R.H.S.

L.H.S.

Chapter 6 Trigonometric Ratios Miscellaneous Exercise

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° 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°

Chapter 6 Trigonometric Ratios Miscellaneous Exercise

Question 13.
If cot θ =  then prove that

Solution :
Given, cot θ = 
cot θ = cot 60°
∴ θ = 60°

Question 14.
Prove 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°) = 
Solution :
4(sin4 30° + cos 60°) – 3(cos2 45° – sin2 90°)

Thus, 4(sin4 30° + cos 60°) – 3(cos2 45° – sin2 90°) = .

Chapter 6 Trigonometric Ratios Miscellaneous Exercise

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 ×  – 6 ×  = 7 – 1 = 6
Thus, 2(cos2 45° + tan2 60°) – 6(sin2 45° – tan2 30°) = 6.

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