Question

Prove the following trigonometric identities.
(i)$\sqrt{\frac{1+sinA}{1-sinA} }=secA+tan A$
(ii)$\sqrt{\frac{1-cosA}{1+cosA} }=cosecA-cot A$

Answer 1

Answer with Step-by-step explanation:

Given :

(i) √(1 + sinA)/(1 − sinA) = sec A + tan A

LHS = √(1 + sinA)/√(1 - sinA)  

= √[(1 + sinA) × √(1 + sinA)] / [√(1 -sinA) × √(1 + sinA) ]  

[By rationalising]

= √(1 + sinA)²/√(1 - sin²A)

[By using an identity , (a + b) (a - b) = a² - b² ]

= (1 + sinA)/√cos²A

[By using the identity, (1- sin²θ) = cos²θ

= (1 + sinA)/cosA

= 1/cosA + sinA/cosA

= secA + tanA  

[By using the identity, secθ = 1/ cosθ & tanθ = sinθ/cosθ]

√(1 + sinA)/(1 − sinA) = sec A + tan A

L.H.S = R.H.S  

Hence Proved..

HOPE THIS ANSWER WILL HELP YOU...

(ii)

Given : √(1-  cosA)/(1 + cosA) = cosec A - cotA

LHS : √(1-  cosA)/(1+ cosA)

= √[(1-  cosA) ×(1-  cosA) /(1+ cosA) × (1-  cosA)]

[By rationalising]

= √(1-  cosA)²/(1 - cos²A)

[By using an identity , (a + b) (a - b) = a² - b²]

= √[(1-  cosA)²/(sin²A)]

[By using  an identity, (1- cos²θ) = sin²θ]

= (1 -  cosA)/(sinA)

= 1/sinA - cosA/sinA

= cosecA - cotA

[By using the identity, cosecθ = 1/sinθ & cotθ = cos θ/sinθ]

√(1-  cosA)/(1+ cosA) = cosec A - cotA

L.H.S = R.H.S  

Hence Proved..

HOPE THIS ANSWER WILL HELP YOU...

Answer 2

Answer:

Refer to the attachment for your answer.