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

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Question

(d) $\sqrt {\frac{{1 - \sin A}}{{1 + \sin A}}} = \sqrt {\frac{{1 - \cos \left( {\frac{\pi }{2} - A} \right)}}{{1 + \cos \left( {\frac{\pi }{2} - A} \r

Vedclass · Accepted Answer

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

Vedclass · Answer

(d) $\sqrt {\frac{{1 - \sin A}}{{1 + \sin A}}} = \sqrt {\frac{{1 - \cos \left( {\frac{\pi }{2} - A} ight)}}{{1 + \cos \left( {\frac{\pi }{2} - A} ight)}}} $

$ = \sqrt {\frac{{2{{\sin }^2}\left( {\frac{\pi }{4} - \frac{A}{2}} ight)}}{{2{{\cos }^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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