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

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Question

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

Vedclass · Accepted Answer

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

Vedclass · Answer

(d) $\frac{{\cos A}}{{1 - \sin A}} = \frac{{\cos A(1 + \sin A)}}{{{{\cos }^2}A}} = \frac{{(1 + \sin A)}}{{\cos A}}$

$ = \frac{{{{\left( {\cos \frac{A}{2} + \sin \frac{A}{2}} ight)}^2}}}{{\left( {\cos \frac{A}{2} + \sin \frac{A}{2}} ight)\,\left( {\cos \frac{A}{2} - \sin \frac{A}{2}} ight)}} $

$= \frac{{\cos \frac{A}{2} + \sin \frac{A}{2}}}{{\cos \frac{A}{2} - \sin \frac{A}{2}}}$

$ = \frac{{1 + 	an \frac{A}{2}}}{{1 - 	an \frac{A}{2}}}$, $\left( {{m{Dividing}}\,{N^r}\,{m{and}}\,{D^r}\,{m{by}}\,\cos \frac{A}{2}} ight)$

$ = 	an \left( {\frac{\pi }{4} + \frac{A}{2}} ight)$.

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

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