You can Download Samacheer Kalvi 10th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

## Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 6 Trigonometry Ex 6.1

**10th Maths Exercise 6.1 Samacheer Kalvi Question 1.**

Prove the following identities.

(i) cot θ + tan θ = sec θ cosec θ

(ii) tan^{4}θ + tan^{2}θ = sec^{4}θ – sec^{2}θ

Solution:

**10th Maths Trigonometry Exercise 6.1 Question 2.**

Prove the following identities

Solution:

**Exercise 6.1 Class 10 Samacheer Kalvi Question 3.**

Prove the following identities

Solution:

**10th Trigonometry Exercise 6.1 Question 4.**

Prove the following identities

(i) sec^{6}θ = tan^{6}θ + 3tan^{2}θ sec^{2}θ + 1

(ii) (sinθ + secθ)^{2} + (cosθ + cosecθ)^{2}

= 1 + (secθ + cosecθ)^{2}

(i) L.H.S = sec^{6}θ = (sec^{2}θ)^{3} = (1 + tan^{2}θ )^{3} = (tan^{2}θ + 1)^{3}

(a + b)^{3} = a^{3} + 3a^{2}b + 3ab^{2} + b^{3}

= (tan^{2}θ)^{3} + 3(tan^{2}θ)^{2} × 1 + 3 × tan^{2}θ × 1^{2} + 1

= tan^{6}θ + 3tan^{2}θ × (sec^{2}θ – 1) + 3tan^{2}θ + 1

= tan^{6}θ + 3tan^{2}θsec^{2}θ – 3tan^{2}θ + 3tan^{2}θ +1

= tan^{6}θ + 3tan^{2}θ sec^{2}θ + 1 = R.H.S

(ii) L.H.S = (sinθ + secθ )^{2} + (cosθ + cosecθ)^{2}

= sin^{2}θ + 2sinθ secθ + sec^{2}θ + cos^{2}θ + 2cosθ cosecθ+ cosec^{2}θ

**Ex 6.1 Class 10 Samacheer Question 5.**

Prove the following identities

Solution:

**10th Maths Exercise 6.1 Question 6.**

Prove the following identities

Solution:

**Trigonometry Exercise 6.1 Question 7.**

Solution:

**10th Maths 6.1 Question 8.**

Solution:

(i) LHS:

**10th Maths Exercise 6.1 In Tamil Question 9.**

(i) If sinθ + cosθ = p and secθ + cosecθ = q then prove that q(p^{2} – 1) = 2p

(ii) If sinθ(1 + sin^{2}θ) = cos^{2}θ, then prove that cos^{6}θ – 4cos^{4}θ + 8cos^{2}θ = 4

Solution:

(ii) Given sinθ(1 + sin^{2}θ) = cos^{2}θ

Sustitute sin^{2}θ = 1 – co^{2}θ and take cos θ = c

squaring (1) on bothsides we get

sin^{2}θ(1 + sin^{2}θ)^{2 }= cos^{4}θ

(1 – c^{2})(1 + 1 – c^{2})^{ }= c^{4
}(1 – c^{2})(2 – c^{2})^{2 }= c^{4}

(1 – c^{2})(4 + c^{4}– 4c^{2}) = c^{4}

4 + c^{4}– 4c^{2 }– 4c^{2} – c^{6} + 4c^{4} = c^{4
} -c^{6 +} 4c^{4} – 8c^{2} = -4

c^{6 }– 4c^{4} + 8c^{2} = -4

ie cos 6θ – 4cos 4θ + 8cos^{2}θ = 4 = RHS

∴ Hence proved

**10th Samacheer Kalvi Maths Trigonometry Question 10.**

Solution: