#### Answer

$(\sec \alpha-\tan \alpha)^2=\frac{1-\sin \alpha}{1+\sin \alpha}$

#### Work Step by Step

Start with the right side:
$\frac{1-\sin \alpha}{1+\sin \alpha}$
Multiply top and bottom by $1-\sin\alpha$:
$=\frac{1-\sin \alpha}{1+\sin \alpha}*\frac{1-\sin\alpha}{1-\sin\alpha}$
$=\frac{1-2\sin\alpha+\sin^2\alpha}{1-\sin^2\alpha}$
$=\frac{1-2\sin\alpha+\sin^2\alpha}{\cos^2\alpha}$
$=\frac{1}{\cos^2\alpha}-\frac{2\sin\alpha}{\cos^2\alpha}+\frac{\sin^2\alpha}{\cos^2\alpha}$
$=\sec^2 \alpha-2*\frac{1}{\cos\alpha}*\frac{\sin\alpha}{\cos\alpha}+\tan^2 \alpha$
$=\sec^2\alpha-2\sec\alpha\tan\alpha+\tan^2 \alpha$
$=(\sec\alpha-\tan\alpha)^2$
Since this equals the left side, the identity has been proven.