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

## Rajasthan Board RBSE Class 10 Maths Chapter 6 Trigonometric Ratios Additional Questions

**Multiple Choice Questions**

Question 1.

It tan θ = 1, then (RBSESolutions.com) cos θ will be :

(A) 1

(B) \(\frac { 1 }{ 2 }\)

(C) \(\frac { \sqrt { 3 } }{ 2 }\)

(D) \(\frac { 1 }{ \sqrt { 2 } }\)

Solution :

tan θ = 1

⇒ tan θ = tan 45°

∴ θ = 45°

cos 45° = \(\frac { 1 }{ \sqrt { 2 } }\)

Hence, option (D) is correct.

Question 2.

If sin θ = cos θ (θ = acute angle), then value of θ will be ;

(A) 30°

(B) 45°

(C) 60°

(D) 75°

Solution :

sin θ = cos θ

⇒ \(\frac { sin\theta }{ cos\theta } \) = 1

⇒ tan θ = 1

⇒ tan θ = tan 45°

∴ θ = 45°

Hence, option (B) is correct.

Question 3.

value of 2 sin^{2} 60° + 3 cot^{2} 30° – tan 45° is.

(A) \(\frac { 2 }{ 19 }\)

(B) \(\frac { 12 }{ 19 }\)

(C) \(\frac { 19 }{ 2 }\)

(D) None of these

Solution :

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

Hence, option (C) is correct.

Question 4.

If in ∆ABC, ∠C is right-angle, (RBSESolutions.com) then value of cos (∠A + ∠B) is :

(A) 0

(B) 1

(C) \(\frac { 1 }{ 2 }\)

(D) \(\frac { \sqrt { 3 } }{ 2 }\) [NCERT Exemplar Problem]

Solution :

Given : ∠C of ∆ABC is right angle.

∵ We know that sum of all the three angles of a triangle is 180°.

∴ ∠A + ∠B + ∠C = 180°

⇒ ∠A + ∠B + 90° = 180°

⇒ ∠A + ∠B = 180° – 90°

⇒ ∠A + ∠B = 90°

Taking cos both sides.

⇒ cos (∠A + ∠B) = cos 90°

⇒ cos(∠A + ∠B) = 0

Hence, option (A) is correct.

Question 5.

If tan θ = √3, then value (RBSESolutions.com) of sin θ is :

(A) \(\frac { 1 }{ \sqrt { 3 } }\)

(B) \(\frac { \sqrt { 3 } }{ 2 }\)

(C) \(\frac { 2 }{ \sqrt { 3 } }\)

(D) 1

Solution :

tan θ = \(\frac { AB }{ BC }\) = √3 = \(\frac { \sqrt { 3 } }{ 1 }\)

Let AB = √3 k

and BC = k

In right angled ∆ABC, by (RBSESolutions.com) pythagorus theorem,

AC^{2} =AB^{2} + BC^{2}

= (√3k)^{2} + (k)^{2}

= 3k^{2} + k^{2} = 4k^{2}

∴ AC = 2k

Now,

Hence, option (B) is correct.

Question 6.

sin (45° + θ) – cos (45° – θ) equals :

(A) 2 cos θ

(B) 0

(C) 2 sinθ

(D) 1 [NCERT Exemplar Problem]

Solution :

sin (45° + θ) – cos (45° – θ)

= sin (45° + θ) – cos [90° – (45° + θ)] [∵ (45° – θ) = {90° – (45° + θ)}]

= sin (45° + θ) – sin (45° + θ) [∵ cos (90° – θ) = sin θ]

= 0

Hence (B) is correct.

Question 7.

Given sin α = \(\frac { 1 }{ 2 }\) and cos β = \(\frac { 1 }{ 2 }\), then (RBSESolutions.com) value of (α + β) is :

(A) 0°

(B) 30°

(C) 60°

(D) 90° [NCERT Exemplar Problem]

Solution :

Given,

sin α = \(\frac { 1 }{ 2 }\) ⇒ sin α = sin 30°

α =30°

and cos β = \(\frac { 1 }{ 2 }\) ⇒ cos β = cos 60°

β = 60°

Now, α + β = 30° + 60° = 90°

Hence, option (D) is correct.

Question 8.

If tan θ = 1, then (RBSESolutions.com) cos θ will be :

(A) 1

(B) \(\frac { 1 }{ 2 }\)

(C) \(\frac { \sqrt { 3 } }{ 2 }\)

(D) \(\frac { 1 }{ \sqrt { 2 } }\)

Solution :

tan θ = 1

⇒ tan θ =tan

∴ θ = 45°

Hence, option (D) is correct.

Question 9.

If sec θ = √2, then (RBSESolutions.com) value of θ will be :

(A) \(\frac { \pi }{ 8 }\)

(B) \(\frac { \pi }{ 6 }\)

(C) \(\frac { \pi }{ 4 }\)

(D) \(\frac { \pi }{ 3 }\)

Solution :

sec θ = √2

sec θ = sec 45° = sec \(\frac { \pi }{ 4 }\)

θ = \(\frac { \pi }{ 4 }\)

Hence, option (C) is correct.

Question 10.

If θ = 30°, then \(\frac { { 1-sin }^{ 2 }2\theta }{ cos2\theta }\) is

(A) 1

(B) -1

(C) 2

(D) -2

Solution :

Hence, (RBSESolutions.com) option (B) is correct.

**Short/Long Answer Type Questions**

Question 1.

Evaluate : cosec^{2} 30° – 3 sec 60°.

Solution :

cosec^{2} 30° – 3 sec 60°

= (2)^{2} – 3(2)

= 4 – 3 × 2 = 4 – 6

= -2

Question 2.

Evaluate : 4 cos^{3} 45° – 3 cos 45°

Solution :

4 cos^{3} 45° – 3 cos 45°

Question 3.

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

(i) sin 60° cos 30° + sin 30° cos 60°

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

Solution :

(i) sin 60° cos 30° + sin 30° cos 60°

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

Question 4.

Prove (RBSESolutions.com) that : \(\frac { { sin60 }^{ \circ } }{ { 1+cos60 }^{ 0 } }\) = tan 30°

Solution :

L.H.S.= \(\frac { { sin60 }^{ \circ } }{ { 1+cos60 }^{ 0 } }\)

R.H.S. = tan 30° = \(\frac { 1 }{ \sqrt { 3 } }\)

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

Question 5.

Evaluate : 3 sin 60° – 4 sin^{3} 60°

Solution :

3 sin 60° – 4 sin^{3} 60°

Putting values of (RBSESolutions.com) trigonometric ratios

Question 6.

If tan 3x = sin 45° cos 45° + sin 30°, then find value of x.

Solution :

tan 3x = sin 45° cos 45° + sin 30°

⇒ tan 3x = \(\frac { 1 }{ \sqrt { 2 } }\) × \(\frac { 1 }{ \sqrt { 2 } }\) + \(\frac { 1 }{ 2 }\)

⇒ tan 3x = \(\frac { 1 }{ 2 }\) + \(\frac { 1 }{ 2 }\)

⇒ tan 3x = 1

⇒ tan 3x = tan 45°

⇒ 3x = 45°

⇒ x = 15°

Question 7.

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

(i) cos(A – B) = cos A cos B + sin A sin B

(ii) sin(A – B) = sin A cos B – cos A sin B

Solution :

(i) A = B = 60° (Given)

LH.S. cos(A – B) = cos(60° – 60°)

= cos 0° = 1

R.HS. cosA cosB + sinA sinB

= cos 60° cos 60° + sin 60° sin 60°

cos^{2} 60° + sin^{2} 60° = 1 [ sin^{2} θ + cos^{2} θ = 1]

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

(ii) sin(A – B) = sin A cos B – cos A sin B

L.H.S. sin(A – B) = sin(60° – 60°)

= sin 0° = 0

R.H.S. sin A cos B – cos A sin B

= sin 60° cos 60° – cos 60° sin 60°

= 0

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

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

## Leave a Reply