Prove the following trigonometric identities:(secA-tanA)^2 = 1-sinA/1+sinA

Question

Prove the following trigonometric identities:

(secA-tanA)² = \({\frac{1-sinA}{1+sinA}}\)

Answer 1

(sec A - tan A)² = \(\Big(\frac{1}{cosA}-\frac{sinA}{cosA}\Big)^2\)

 = \(\frac{(1-sinA)^2}{cos^2A}\)  

=  \(\frac{(1-sinA)^2}{1-sin^2A}\)  

= \(\frac{(1-sinA)^2}{(1-sinA)(1+sinA)}\)

=  \(\frac{1-sinA}{1+sinA}\)  

Hence Proved.

Answer 2

sec A = 1/cos A

tan A = sin A/ cos A

Thus,

L.H.S=(1/cosA - sin A/cos A)²

=(1-sin A)²/cos² A

=(1-sin A)(1- sin A)/(1-sin² A)                      By a²-b²=(a+b)(a-b) and sin² A + cos² A=1

=(1-sin A)/(1+sin A)=R.H.S(corrected)

Hence, proved