**RBSE Solutions for Class 10 Maths Chapter 6 Trigonometric Ratios** Ex 6.1 is part of RBSE Solutions for Class 10 Maths. Here we have given Rajasthan Board RBSE Class 10 Maths Chapter 6 Trigonometric Ratios Ex 6.1.

## Rajasthan Board RBSE Class 10 Maths Chapter 6 Trigonometric Ratios Ex 6.1

Find the value of the (RBSESolutions.com) following : (Q. 1 to 10)

Question 1.

2 sin 45° cos 45°

Solution :

2 sin 45° cos 45° = 2 × ×

= 2 × = 1

Question 2.

cos 45° cos 60° – sin 45° sin 60°.

Solution :

cos 45° cos 60° – sin 45° sin 60°

Question 3.

sin^{2} 30° + 2 cos^{2} 45° + 3 tan^{2} 60°

Solution :

sin^{2} 30° + 2 cos^{2} 45° + 3 tan^{2} 60°

Putting value of (RBSESolutions.com) trigonometric ratios,

Question 4.

3 sin 60° – 4 sin^{3} 60°.

Solution :

3 sin 60° – 4 sin^{3} 60°

Putting value of trigonometric ratios

Question 5.

Solution :

Question 6.

4 cot^{2} 45° – sec^{2} 60° + sin^{2} 60° + cos^{2} 90°.

Solution :

4 cot^{2} 45° – sec^{2} 60° + sin^{2} 60° + cos^{2} 90°

Question 7.

Solution :

Question 8.

Solution :

Question 9.

Solution :

Question 10.

Solution :

Question 11.

Find the value (RBSESolutions.com) of x in the following :

(i) cos x = cos 60° cos 30° + sin 60° sin 30°

(ii) sin 2x = sin 60° cos 30° – cos 60° sin 30°

(iii) √3 tan2x = sin 30° + sin 45° cos 45° + 2 sin 90°.

Solution :

(i) cos x = cos 60° cos 30° + sin 60° sin 30°

cos x = cos 30°

x = 30°

(ii) sin 2x = sin 60° cos 30° – cos 60° sin 30°

⇒ sin 2x = = sin 30°

⇒ 2x = 30° ⇒ x = 15°

(iii) √3 tan2x = sin 30° + sin 45° cos 45° + 2 sin 90°

√3 tan2x = 3

⇒ tan 2x =

⇒ tan 2x = √3

⇒ tan 2x = tan 60°

⇒ 2x = 60°

⇒ x = 30°

Prove that : (Q. 12 to Q. 20)

Question 12.

Solution :

L.H.S.

= R.H.S.

Question 13.

4 cot^{2} 45° – sec^{2} 60° – sin^{2} 30° = –

Solution :

4 cot^{2} 45° – sec^{2} 60° – sin^{2} 30° = –

L.H.S. = 4 cot^{2} 45° – sec^{2} 60° – sin^{2} 30°

= 4(1)^{2} – (2)^{2} – { \left( \frac { 1 }{ 2 } \right) }^{ 2 }

= 4 – 4 – = –

= R.H.S.

Hence, L.H.S. = R.H.S.

Question 14.

4 sin 30° sin^{2} 60° + 3 cos 60° tan 45° = 2 sec^{2} 45° – cosec^{2} 90°

Solution :

L.H.S. = 4 sin 30° sin^{2} 60° + 3 cos 60° tan 45°

Putting values of (RBSESolutions.com) trigonometric ratios,

R.H.S. = 2 sec^{2} 45° – cosec^{2} 90°

Putting values of trigonometric ratios,

= 2(√2)^{2} – (1)^{2} = 4 – 1 = 3

∴ L.H.S. = R.H.S.

Question 15.

cosec^{2} 45° sec^{2} 30° sin^{3} 90° cos 60° =

Solution :

L.H.S. = cosec^{2} 45° sec^{2} 30° sin^{3} 90° cos 60°

∴ L.H.S. = R.H.S.

Question 16.

Solution :

L.H.S.

Thus, L.H.S. = R.H.S.

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°)

= 7 – 1 = 6

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

Question 18.

(sec^{2} 30° + cosec^{2} 45°) (2 cos 60°+ sin 90° + tan 45°) = 10

Solution :

(sec^{2} 30° + cosec^{2} 45°) (2 cos 60°+ sin 90° + tan 45°)

Thus, (sec^{2} 30° + cosec^{2} 45°) (2 cos 60°+ sin 90° + tan 45°) = 10

Question 19.

(1 – sin 45° + sin 30°) (1 + cos 45° + cos 60°) =

Solution :

(1 – sin 45° + sin 30°) (1 + cos 45° + cos 60°)

Thus, (1 – sin 45° + sin 30°) (1 + cos 45° + cos 60°) =

Question 20.

cos^{2} 0° – 2 cot^{2} 30° + 3 cosec^{2} 90° = 2(sec^{2} 45° – tan^{2} 60°)

Solution :

L.H.S. = cos^{2} 0° – 2 cot^{2} 30° + 3 cosec^{2} 90°

= (1)^{2} – 2(√3)^{2} + 3(1)^{2}

= 1 – 2 × 3 + 3

= 1 – 6 + 3 = -2

R.H.S. = 2(sec^{2} 45° – tan^{2} 60°)

= 2[(√2)^{2} – (√3)^{2}]

= 2(2 – 3) = 2 × (-1) = -2

Thus, L.H.S. = R.H.S.

Question 21.

If x = 30°, then (RBSESolutions.com) prove that

(i) sin 3x = 3 sin x – 4 sin^{3} x

(iv) cos 3x = 4 cos^{3} x – 3 cos x

Solution :

(i) sin 3x = 3 sin x – 4 sin^{3} x

L.H.S. = sin 3x

= sin 3 × 30° [∵ Given x = 30°]

= sin 90°

= 1

R.H.S. = 3 sin x – 4 sin^{3} x

= 3sin 30° – 4 sin^{3} 30°

Thus, L.H.S. = R.H.S.

(ii) tan 2x =

L.H.S. = tan 2x

= tan 2 × 30° [∵ Given x = 30°]

= tan 60° = √3

R.H.S. =

∴ L.H.S. = R.H.S.

(iii)

∴ L.H.S. = R.H.S.

(iv) cos 3x = 4 cos^{3} x – 3 cos x

L.H.S. = cos 3x = cos 3 × 30°

= cos 90°

= 0 [∵ Given x = 30°]

R.H.S. = 4 cos^{3} x – 3 cos x = 4 cos^{3} 30° – 3 cos 30°

L.H.S. = R.H.S.

Question 22.

If A = 60° and B = 30° then (RBSESolutions.com) prove that :

cot (A – B) =

Solution :

cot (A – B) =

∴ A = 60°, B = 30°

L.H.S. = cot(A-B) = cot(60° – 30°)

= cot 30° = √3

R.H.S. =

L.H.S. = R.H.S.

We hope the given RBSE Solutions for Class 10 Maths Chapter 6 Trigonometric Ratios Ex 6.1 will help you. If you have any query regarding Rajasthan Board RBSE Class 10 Maths Chapter 6 Trigonometric Ratios Exercise 6.1, drop a comment below and we will get back to you at the earliest.

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