# RD Sharma Class 10 Ex 6.2 Solutions Chapter 6 Trigonometric Identities

In this chapter, we provide RD Sharma Class 10 Ex 6.2 Solutions Chapter 6 Trigonometric Identities for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Class 10 Ex 6.2 Solutions Chapter 6 Trigonometric Identities pdf, Now you will get step by step solution to each question.

# Chapter 6: Trigonometric Identities Exercise – 6.2

### Question: 1

If cos θ = 4/5, find all other trigonometric ratios of angle θ.

### Solution:

We have: sin θ = 1 – cos2θ

= 3/5

Therefore, sin θ = 3/5

### Question: 2

If sin θ = 1/√2, find all other trigonometric ratios of angle θ.

We have,

=1/1

=1

### Solution:

We know that

Substituting it in equation (1) we get

### Solution:

We know that

sec θ = 5/4

= 1/(5/4)

= 4/5

= cos θ

Therefore, We get

= 12/13

i. e. We get

= 25/1

= 25.

### Solution:

= cos θ = sin θ × cot θ

= 1/2

Therefore, on substituting we get

= 3/5.

### Solution:

We know that

= 1.

= 1/1

= 1

On substituting we get:

= 4/2

= 2

### Solution:

= 2

cos θ = cot θ. sin θ

On substituting we get:

= 21/8

### Solution:

On substituting we get

= 40/4

= 10

### Question: 10

If √3 tan θ = sin θ, find the value of sin2θ – cos2θ.

= sin2θ – cos2θ

= 1/3

= 12/13

= 9/3

= 3

### Question: 12

If sin θ + cos θ =√2 cos(90°- θ), find cot θ.

### Solution:

= sin θ + cos θ = √2 sin θ [cos (90 – θ) = sin θ]

⇒ cos θ = √2 sin θ – sin θ

⇒ cos θ = sin θ (√2 – 1)

Divide both sides with sin θ we get

= √2 – 1.

### Question: 13

If 2 sinθ − cos2θ = 2, then find the value of θ.

### Solution:

2 sinθ − cos2 θ = 2

⇒ 2 sin2 θ − (1 − sin2 θ) = 2

⇒ 2 sin2 θ − 1 + sin2 θ = 2

⇒ 3 sin2 θ = 3

⇒ sin2 θ = 1

⇒ sin θ = 1

⇒ sin θ =  sin 90°

⇒ θ = 90°

### Question: 14

If √3 tan θ – 1 = 0, find the value of sin2 θ – cos2 θ.

### Solution:

√3 tan θ = tan 30° θ = 30°

Now, sin2θ – cos2 θ = sin2 (30°) – cos2 (30°)

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