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

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Question

(C) We know that $1 = \sec^2 A - 	an^2 A$. Substituting this in the numerator:$\frac{{	an A + \sec A - (\sec^2 A - 	an^2 A)}}{{	an A - \sec A + 1}

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) We know that $1 = \sec^2 A - 	an^2 A$. Substituting this in the numerator:$\frac{{	an A + \sec A - (\sec^2 A - 	an^2 A)}}{{	an A - \sec A + 1}}$$= \frac{{(	an A + \sec A) - (\sec A - 	an A)(\sec A + 	an A)}}{{	an A - \sec A + 1}}$$= \frac{{(	an A + \sec A)(1 - (\sec A - 	an A))}}{{	an A - \sec A + 1}}$$= \frac{{(	an A + \sec A)(1 - \sec A + 	an A)}}{{	an A - \sec A + 1}}$Since the term $(1 - \sec A + 	an A)$ is common in the numerator and denominator,they cancel out:$= 	an A + \sec A$$= \frac{{\sin A}}{{\cos A}} + \frac{1}{{\cos A}}$$= \frac{{1 + \sin A}}{{\cos A}}$Thus,the correct option is $C$.

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

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