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.

RD Sharma Solutions for Class 10 Chapter 6 Trigonometric Ratios Ex 6.2 Download PDF

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.

Solution: 

We have,

=1/1

=1

Question: 3

Solution: 

We know that

 Substituting it in equation (1) we get

Question: 4

Solution: 

We know that

sec θ = 5/4

= 1/(5/4)

= 4/5

= cos θ

Therefore, We get

Question: 5

Solution: 

= 12/13

 i. e. We get

= 25/1

= 25.

Question: 6

Solution: 

= cos θ = sin θ × cot θ

= 1/2

Therefore, on substituting we get

= 3/5.

Question: 7

Solution: 

We know that

= 1.

= 1/1

= 1

On substituting we get:

= 4/2

= 2 

Question: 8

Solution: 

= 2

cos θ = cot θ. sin θ

On substituting we get:

= 21/8

Question: 9

Solution:

On substituting we get

= 40/4

= 10

Question: 10

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

Solution: 

= sin²θ – cos²θ

= 1/3

Question: 11

Solution: 

= 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 sin² θ − cos²θ = 2, then find the value of θ. 

Solution: 

2 sin² θ − 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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