Question
Prove the following identitie:$\frac{\cos A}{1+\sin A}+\tan A=\sec A$
✓
Answer
$\frac{\cos A}{1+\sin A}+\tan A$
$=\frac{\cos A}{1+\sin A}+\frac{\sin A}{\cos A}$
$=\frac{\cos ^2 A+\sin A+\sin ^2 A}{(1+\sin ) \cos A}$
$=\frac{1+\sin A}{(1+\sin A) \cos A}$
$=\frac{\cos ^3 A+\cos A \sin A-\sin ^2 A}{\cos ^2 A-\sin A \cos A}$
$=\frac{1}{\cos A}$
$=\sec A$
$=\frac{\cos A}{1+\sin A}+\frac{\sin A}{\cos A}$
$=\frac{\cos ^2 A+\sin A+\sin ^2 A}{(1+\sin ) \cos A}$
$=\frac{1+\sin A}{(1+\sin A) \cos A}$
$=\frac{\cos ^3 A+\cos A \sin A-\sin ^2 A}{\cos ^2 A-\sin A \cos A}$
$=\frac{1}{\cos A}$
$=\sec A$
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