frac{{1 + sin A - cos A}}{{1 + sin A + cos A} | vedclass

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Question

$\frac{{1 + \sin A - \cos A}}{{1 + \sin A + \cos A}}$

$ = \frac{{2\,{{\sin }^2}\frac{A}{2} + 2\,\sin \frac{A}{2}\cos \frac{A}{2}}}{{2\,{{\cos }^2}\

Vedclass · Accepted Answer

$	an \frac{A}{2}$

Vedclass · Answer

$\frac{{1 + \sin A - \cos A}}{{1 + \sin A + \cos A}}$

$ = \frac{{2\,{{\sin }^2}\frac{A}{2} + 2\,\sin \frac{A}{2}\cos \frac{A}{2}}}{{2\,{{\cos }^2}\frac{A}{2} + 2\,\sin \frac{A}{2}\cos \frac{A}{2}}}$

$ = \frac{{2\,\,\sin \frac{A}{2}\,\left( {\sin \frac{A}{2} + \cos \frac{A}{2}} ight)}}{{2\,\,\cos \frac{A}{2}\,\left( {\cos \frac{A}{2} + \sin \frac{A}{2}} ight)}}$=$	an \frac{A}{2}$.

ट्रिक :  $A = {60^o}$ रखने पर,

$\frac{{1 + (\sqrt 3 /2) - (1/2)}}{{1 + (\sqrt 3 /2) + (1/2)}} = \frac{{1 + \sqrt 3 }}{{3 + \sqrt 3 }} = \frac{1}{{\sqrt 3 }}$

जो कि विकल्प $(c)$ है अर्थात् $	an \frac{{{{60}^o}}}{2} = \frac{1}{{\sqrt 3 }}$ है।

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

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