Question
Prove the following identitie:$\frac{\cos A}{1+\sin A}+\tan A=\sec A$

Answer

$\frac{\cos A}{1+\sin A}+\tan A $
$ =\frac{\cos A}{1+\sin A}+\frac{\sin A}{\cos A} $
$ =\frac{\cos ^2 A+\sin A+\sin ^2 A}{(1+\sin ) \cos A} $
$ =\frac{1+\sin A}{(1+\sin A) \cos A} $
$ =\frac{\cos ^3 A+\cos A \sin A-\sin ^2 A}{\cos ^2 A-\sin A \cos A} $
$=\frac{1}{\cos A} $
$=\sec A$

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