Question
Prove.
$\frac{\cos A}{1-\sin A}=\sec A+\tan A$
$\frac{\cos A}{1-\sin A}=\sec A+\tan A$
✓
Answer
$\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) $
$=\frac{\cos ^2 A}{\cos A(1-\sin A)}=\frac{\cos A}{1-\sin A}=\text { LHS }$
$\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) $
$=\frac{\cos ^2 A}{\cos A(1-\sin A)}=\frac{\cos A}{1-\sin A}=\text { LHS }$
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