Question
Prove that $\frac{\tan A+\sec A-1}{\tan A-\sec A+1}=\frac{1+\sin A}{\cos A}$.

Answer

$\text { LHS }=\frac{\tan A+\sec A-1}{\tan A-\sec A+1} $
$ =\frac{(\tan A+\sec A)-\left(\sec ^2 A-\tan ^2 A\right)}{(\tan A-\sec A)+1} $
$ =\frac{(\tan A+\sec A)(1-\sec A+\tan A)}{\tan A-\sec A+1}$
$= \tan A + \sec A$
$=\frac{\sin A}{\cos A}+\frac{1}{\cos A}=\frac{1+\sin A}{\cos A}$
$= RHS$
Hence proved.

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