Question
Prove that
$\frac{\cos A}{1+\sin A}=\sec A-\tan A$
$\frac{\cos A}{1+\sin A}=\sec A-\tan A$
✓
Answer
$\begin{aligned} & \frac{\cos A}{1+\sin A} \\ = & \frac{\cos A}{1+\sin A} \times \frac{1-\sin A}{1-\sin A} \\ = & \frac{\cos A(1-\sin A)}{1-\sin ^2 A} \\ = & \frac{\cos A(1-\sin A)}{\cos ^2 A} \\ = & \frac{1-\sin A}{\cos A} \\ = & \sec A-\tan A\end{aligned}$
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