Question
Prove the following identity:
$
\frac{\cos A}{1+\sin A}=\sec A-\tan A
$
$
\frac{\cos A}{1+\sin A}=\sec A-\tan A
$
✓
Answer
$
\begin{aligned}
& \text { LHS }=\frac{\cos A}{1+\sin A} \\
& \text { RHS }=\sec A-\tan A \\
& =\frac{1}{\cos A}-\frac{\sin A}{\cos A}=\frac{1-\sin A}{\cos A} \\
& =\frac{1-\sin A}{\cos A}\left(\frac{1+\sin A}{1+\sin A}\right)=\left(\frac{1-\sin ^2 A}{\cos A(1+\sin A)}\right) \\
& =\frac{\cos ^2 A}{\cos A(1+\sin A)}=\frac{\cos A}{(1+\sin A)}=\mathrm{LHS}
\end{aligned}
$
\begin{aligned}
& \text { LHS }=\frac{\cos A}{1+\sin A} \\
& \text { RHS }=\sec A-\tan A \\
& =\frac{1}{\cos A}-\frac{\sin A}{\cos A}=\frac{1-\sin A}{\cos A} \\
& =\frac{1-\sin A}{\cos A}\left(\frac{1+\sin A}{1+\sin A}\right)=\left(\frac{1-\sin ^2 A}{\cos A(1+\sin A)}\right) \\
& =\frac{\cos ^2 A}{\cos A(1+\sin A)}=\frac{\cos A}{(1+\sin A)}=\mathrm{LHS}
\end{aligned}
$
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