**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 of tan^{2} 60° is :

(A) 3

(B)

(C) 1

(D) ∞

Solution :

tan^{2} 60° = (√3)^{2} = 3

(A) is correct.

Question 2.

The value of 2 sin^{2} 60° cos 60° is :

(A)

(B)

(C)

(D)

Solution :

2 sin^{2} 60° cos 60°

Hence (C) is correct.

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 cos^{2} = 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 cos^{2} 30° – 1

Solution :

R.H.S.

L.H.S. = cos 60° =

∴ 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° sin 60° sin 90° = √2

Solution :

L.H.S. = 4 tan 30° sin 45° sin 60° sin 90°

= R.H.S.

Question 11.

Evaluate : sin^{2} 60° cot^{2} 60°.

solution :

sin^{2} 60° cot^{2} 60°

Question 12.

Evaluate :

4 cos^{3} 30° – 3 cos 30°.

Solution :

4 cos^{3} 30° – 3 cos 30°

Question 13.

If cot θ = then prove that

Solution :

Given, cot θ =

cot θ = cot 60°

∴ θ = 60°

Question 14.

Prove that 3(tan^{2} 30° + cot^{2} 30°) – 8(sin^{2} 45° + cos^{2} 30°) = 0

Solution :

3(tan^{2} 30° + cot^{2} 30°) – 8(sin^{2} 45° + cos^{2} 30°)

= 10 – 10 = 0

Thus, 3(tan^{2} 30° + cot^{2} 30°) – 8(sin^{2} 45° + cos^{2} 30°) = 0

Question 15.

4(sin^{4} 30° + cos 60°) – 3(cos^{2} 45° – sin^{2} 90°) =

Solution :

4(sin^{4} 30° + cos 60°) – 3(cos^{2} 45° – sin^{2} 90°)

Thus, 4(sin^{4} 30° + cos 60°) – 3(cos^{2} 45° – sin^{2} 90°) =

Question 16.

Solution :

Question 17.

2(cos^{2} 45° + tan^{2} 60°) – 6(sin^{2} 45° – tan^{2} 30°) = 6

Solution :

2(cos^{2} 45° + tan^{2} 60°) – 6(sin^{2} 45° – tan^{2} 30°)

= 2 × – 6 × = 7 – 1 = 6

Thus, 2(cos^{2} 45° + tan^{2} 60°) – 6(sin^{2} 45° – tan^{2} 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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