frac{{tan A + sec A - 1}}{{tan A - sec A + 1} | vedclass

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Question

(c) $\frac{{	an A + \sec A - 1}}{{	an A - \sec A + 1}}$

$ = \frac{{\sin A - \cos A + 1}}{{\sin A - 1 + \cos A}} $

$= \frac{{\sin A + (1 - \cos

Vedclass · Accepted Answer

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

Vedclass · Answer

(c) $\frac{{	an A + \sec A - 1}}{{	an A - \sec A + 1}}$

$ = \frac{{\sin A - \cos A + 1}}{{\sin A - 1 + \cos A}} $

$= \frac{{\sin A + (1 - \cos A)}}{{\sin A - (1 - \cos A)}}$

$ = \frac{{2\sin \frac{A}{2}\cos \frac{A}{2} + 2{{\sin }^2}\frac{A}{2}}}{{2\sin \frac{A}{2}\cos \frac{A}{2} - 2{{\sin }^2}\frac{A}{2}}}$

$ = \frac{{\cos \frac{A}{2} + \sin \frac{A}{2}}}{{\cos \frac{A}{2} - \sin \frac{A}{2}}} $

$= \frac{{{{\left( {\cos \frac{A}{2} + \sin \frac{A}{2}} ight)}^2}}}{{{{\cos }^2}\frac{A}{2} - {{\sin }^2}\frac{A}{2}}}$

$ = \frac{{1 + \sin A}}{{\cos A}}$.

$\frac{{\tan A + \sec A - 1}}{{\tan A - \sec A + 1}} = $

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