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

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

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

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