Samacheer Kalvi 10th Maths Solutions Chapter 6 Trigonometry Ex 6.1

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

Question 1.
Prove the following identities.
(i) cot θ + tan θ = sec θ cosec θ
(ii) tan4θ + tan2θ = sec4θ – sec2θ
Solution:
Samacheer Kalvi 10th Maths Chapter 6 Trigonometry Ex 6.1 1

Samacheer Kalvi 10th Maths Solutions Chapter 6 Trigonometry Ex 6.1

Question 2.
Prove the following identities
Samacheer Kalvi 10th Maths Chapter 6 Trigonometry Ex 6.1 2
Solution:
Samacheer Kalvi 10th Maths Chapter 6 Trigonometry Ex 6.1 50
Samacheer Kalvi 10th Maths Chapter 6 Trigonometry Ex 6.1 3

Question 3.
Prove the following identities
Samacheer Kalvi 10th Maths Chapter 6 Trigonometry Ex 6.1 4
Solution:
Samacheer Kalvi 10th Maths Chapter 6 Trigonometry Ex 6.1 5
Samacheer Kalvi 10th Maths Chapter 6 Trigonometry Ex 6.1 6

Question 4.
Prove the following identities
(i) sec6θ = tan6θ + 3tan2θ sec2θ + 1
(ii) (sinθ + secθ)2 + (cosθ + cosecθ)2
= 1 + (secθ + cosecθ)2
(i) L.H.S = sec6θ = (sec2θ)3 = (1 + tan2θ )3 = (tan2θ + 1)3
(a + b)3 = a3 + 3a2b + 3ab2 + b3
= (tan2θ)3 + 3(tan2θ)2 × 1 + 3 × tan2θ × 12 + 1
= tan6θ + 3tan2θ × (sec2θ – 1) + 3tan2θ + 1
= tan6θ + 3tan2θsec2θ – 3tan2θ + 3tan2θ +1
= tan6θ + 3tan2θ sec2θ + 1 = R.H.S

(ii) L.H.S = (sinθ + secθ )2 + (cosθ + cosecθ)2
= sin2θ + 2sinθ secθ + sec2θ + cos2θ + 2cosθ cosecθ+ cosec2θ
Samacheer Kalvi 10th Maths Chapter 6 Trigonometry Ex 6.1 7

Samacheer Kalvi 10th Maths Solutions Chapter 6 Trigonometry Ex 6.1

Question 5.
Prove the following identities
Samacheer Kalvi 10th Maths Chapter 6 Trigonometry Ex 6.1 8
Solution:
Samacheer Kalvi 10th Maths Chapter 6 Trigonometry Ex 6.1 9
Samacheer Kalvi 10th Maths Chapter 6 Trigonometry Ex 6.1 10
Samacheer Kalvi 10th Maths Chapter 6 Trigonometry Ex 6.1 11

Question 6.
Prove the following identities
Samacheer Kalvi 10th Maths Chapter 6 Trigonometry Ex 6.1 12
Solution:
Samacheer Kalvi 10th Maths Chapter 6 Trigonometry Ex 6.1 13
Samacheer Kalvi 10th Maths Chapter 6 Trigonometry Ex 6.1 14
Samacheer Kalvi 10th Maths Chapter 6 Trigonometry Ex 6.1 15

Question 7.
Samacheer Kalvi 10th Maths Chapter 6 Trigonometry Ex 6.1 16
Solution:
Samacheer Kalvi 10th Maths Chapter 6 Trigonometry Ex 6.1 17
Samacheer Kalvi 10th Maths Chapter 6 Trigonometry Ex 6.1 18

Samacheer Kalvi 10th Maths Solutions Chapter 6 Trigonometry Ex 6.1

Question 8.
Samacheer Kalvi 10th Maths Chapter 6 Trigonometry Ex 6.1 19
Solution:
(i) LHS:
Samacheer Kalvi 10th Maths Chapter 6 Trigonometry Ex 6.1 20
Samacheer Kalvi 10th Maths Chapter 6 Trigonometry Ex 6.1 21
Samacheer Kalvi 10th Maths Chapter 6 Trigonometry Ex 6.1 22
Samacheer Kalvi 10th Maths Chapter 6 Trigonometry Ex 6.1 23

Question 9.
(i) If sinθ + cosθ = p and secθ + cosecθ = q then prove that q(p2 – 1) = 2p
(ii) If sinθ(1 + sin2θ) = cos2θ, then prove that cos6θ – 4cos4θ + 8cos2θ = 4
Solution:
Samacheer Kalvi 10th Maths Chapter 6 Trigonometry Ex 6.1 24
(ii) Given sinθ(1 + sin2θ) = cos2θ
Sustitute sin2θ = 1 – co2θ and take cos θ = c
squaring (1) on bothsides we get
sin2θ(1 + sin2θ)2 = cos4θ
(1 – c2)(1 + 1 – c2) = c4
(1 – c2)(2 – c2)2 = c4
(1 – c2)(4 + c4– 4c2) = c4
4 + c4– 4c2 – 4c2 – c6 + 4c4 = c4
 -c6 + 4c4 – 8c2 = -4
c6 – 4c4 + 8c2 = -4
ie cos 6θ – 4cos 4θ + 8cos2θ = 4 = RHS
∴ Hence proved

Samacheer Kalvi 10th Maths Solutions Chapter 6 Trigonometry Ex 6.1

Question 10.
Samacheer Kalvi 10th Maths Chapter 6 Trigonometry Ex 6.1 26
Solution:
Samacheer Kalvi 10th Maths Chapter 6 Trigonometry Ex 6.1 27

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