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

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(D) We have $\sqrt{\frac{1 - \sin A}{1 + \sin A}}$.Using the identity $\sin A = \cos \left( \frac{\pi}{2} - A \right)$,we get $\sqrt{\frac{1 - \cos \l

Vedclass · Accepted Answer

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

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(D) We have $\sqrt{\frac{1 - \sin A}{1 + \sin A}}$.Using the identity $\sin A = \cos \left( \frac{\pi}{2} - A ight)$,we get $\sqrt{\frac{1 - \cos \left( \frac{\pi}{2} - A ight)}{1 + \cos \left( \frac{\pi}{2} - A ight)}}$.Applying the half-angle formulas $1 - \cos 	heta = 2 \sin^2 \left( \frac{	heta}{2} ight)$ and $1 + \cos 	heta = 2 \cos^2 \left( \frac{	heta}{2} ight)$,we have:$\sqrt{\frac{2 \sin^2 \left( \frac{\pi}{4} - \frac{A}{2} ight)}{2 \cos^2 \left( \frac{\pi}{4} - \frac{A}{2} ight)}} = \sqrt{	an^2 \left( \frac{\pi}{4} - \frac{A}{2} ight)} = 	an \left( \frac{\pi}{4} - \frac{A}{2} ight)$.

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

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