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