sqrt {frac{{1 - sin A}}{{1 + sin A}}} = | vedclass

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(D) We know that $1 - \sin A = 1 - \cos \left( {\frac{\pi }{2} - A} \right) = 2\sin^2 \left( {\frac{\pi }{4} - \frac{A}{2}} \right)$ and $1 + \sin A =

Vedclass · Accepted Answer

$	an \left( {\frac{\pi }{4} - \frac{A}{2}} ight)$

Vedclass · Answer

(D) We know that $1 - \sin A = 1 - \cos \left( {\frac{\pi }{2} - A} ight) = 2\sin^2 \left( {\frac{\pi }{4} - \frac{A}{2}} ight)$ and $1 + \sin A = 1 + \cos \left( {\frac{\pi }{2} - A} ight) = 2\cos^2 \left( {\frac{\pi }{4} - \frac{A}{2}} ight)$.Substituting these into the expression:$\sqrt {\frac{{1 - \sin A}}{{1 + \sin A}}} = \sqrt {\frac{{2\sin^2 \left( {\frac{\pi }{4} - \frac{A}{2}} ight)}}{{2\cos^2 \left( {\frac{\pi }{4} - \frac{A}{2}} ight)}}} = \sqrt {	an^2 \left( {\frac{\pi }{4} - \frac{A}{2}} ight)} = 	an \left( {\frac{\pi }{4} - \frac{A}{2}} ight)$.

$\sqrt {\frac{{1 - \sin A}}{{1 + \sin A}}} = $

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