Trigonometric ratios, in terms of the value of one of them, are the main focus of this exercise. The RD Sharma Solutions Class 10 is a one-stop solution for students to develop good problem-solving skills in various chapters of RD Sharma. All the solutions are prepared by expert teams at BYJU’S following the CBSE patterns. Students can download **RD Sharma Solutions for Class 10 Maths Chapter 6 Trigonometric Identities Exercise 6.2 **PDF below for complete assistance.

## ML Aggarwal Solutions for Class 6 Maths Chapter 2 – Whole Numbers

### Access RD Sharma Solutions for Class 10 Chapter 6 Trigonometric Identities Exercise 6.2

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

**Solution: **

We have,

cos θ = 4/5

And we know that,

sin θ = √(1 – cos^{2 }θ)

⇒ sin θ = √(1 – (4/5)^{2})

= √(1 – (16/25))

= √[(25 – 16)/25]

= √(9/25)

= 3/5

∴ sin θ = 3/5

Since, cosec θ = 1/ sin θ

= 1/ (3/5)

⇒ cosec θ = 5/3

And, sec θ = 1/ cos θ

= 1/ (4/5)

⇒ cosec θ = 5/4

Now,

tan θ = sin θ/ cos θ

= (3/5)/ (4/5)

⇒ tan θ = 3/4

And, cot θ = 1/ tan θ

= 1/ (3/4)

⇒ cot θ = 4/3

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

**Solution: **

We have,

sin θ = 1/√2

And we know that,

cos θ = √(1 – sin^{2 }θ)

⇒ cos θ = √(1 – (1/√2)^{2})

= √(1 – (1/2))

= √[(2 – 1)/2]

= √(1/2)

= 1/√2

∴ cos θ = 1/√2

Since, cosec θ = 1/ sin θ

= 1/ (1/√2)

⇒ cosec θ = √2

And, sec θ = 1/ cos θ

= 1/ (1/√2)

⇒ sec θ = √2

Now,

tan θ = sin θ/ cos θ

= (1/√2)/ (1/√2)

⇒ tan θ = 1

And, cot θ = 1/ tan θ

= 1/ (1)

⇒ cot θ = 1

**3. **

**Solution: **

Given,

tan θ = 1/√2

By using sec^{2 }θ − tan^{2 }θ = 1,

**4. **

**Solution: **

Given,

tan θ = 3/4

By using sec^{2 }θ − tan^{2 }θ = 1,

** **

** sec **θ = 5/4

Since, sec θ = 1/ cos θ

⇒ cos θ = 1/ sec θ

= 1/ (5/4)

= 4/5

So,

**5. **

**Solution: **

Given, tan θ = 12/5

Since, cot θ = 1/ tan θ = 1/ (12/5) = 5/12

Now, by using cosec^{2 }θ − cot^{2 }θ = 1

cosec θ = √(1 + cot^{2 }θ)

= √(1 + (5/12)^{2 })

= √(1 + 25/144)

= √(169/ 25)

⇒ cosec θ = 13/5

Now, we know that

sin θ = 1/ cosec θ

= 1/ (13/5)

⇒ sin θ = 5/13

Putting value of sin θ in the expression we have,

= 25/ 1

= 25

**6. **

**Solution: **

**Given, **

cot θ = 1/√3

Using cosec^{2 }θ − cot^{2 }θ = 1, we can find cosec θ

cosec θ = √(1 + cot^{2} θ)

= √(1 + (1/√3)^{2})

= √(1 + (1/3)) = √((3 + 1)/3)

= √(4/3)

⇒ cosec θ = 2/√3

So, sin θ = 1/ cosec θ = 1/ (2/√3)

⇒ sin θ = √3/2

And, we know that

cos θ = √(1 – sin^{2} θ)

= √(1 – (√3/2)^{2})

= √(1 – (3/4))

** = **√((4 – 3)/4)

= √(1/4)

⇒ cos θ = 1/2

Now, using cos θ and sin θ in the expression, we have

= 3/5

**7. **

**Solution: **

Given,

cosec A = √2

Using cosec^{2 }A − cot^{2 }A = 1, we find cot A

= 4/2

= 2

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