Question

Prove the following trigonometric identities. $(secA-tanA)^{2} =\frac{1-sinA}{1+sinA}$

Answer 1

Answer with Step-by-step explanation:

Given :  

(sec A – tan A)² = (1− sinA)/(1 + sinA)

LHS : (sec A – tan A)²

= [1/cosA − sinA/cosA]²

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

= [ (1− sinA)/cosA]²

[By taking LCM]

= (1 - sinA)²/( cos²A)

= (1− sinA)² / (1− sin²A)

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

= (1− sinA)²/(1+ sinA)(1− sinA)

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

= (1− sinA) (1 - sinA)/(1+ sinA)(1− sinA)

=  (1− sinA)/(1 + sinA)

(sec A – tan A)² = (1− sinA)/(1 + sinA)

L.H.S = R.H.S  

Hence Proved..

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Answer 2

Answer:

Refer to the attachment for your answer.