If

If $-\frac{\pi}{2}

$\sqrt{\frac{1-\sin x}{1+\sin x}}=\sqrt{\frac{\sin ^{2} \frac{x}{2}+\cos ^{2} \frac{x}{2}-2 \sin \frac{x}{2} \cos \frac{x}{2}}{\sin ^{2} \frac{x}{2}+\cos ^{2} \frac{x}{2}+2 \sin \frac{x}{2} \cos \frac{x}{2}}}$

(using identics : $-\sin ^{2} \theta+\cos ^{2} \theta-1 \sin 2 \theta=2 \sin \theta \cos \theta$ )

$=\sqrt{\frac{\left(\sin \frac{x}{2}-\cos \frac{x}{2}\right)^{2}}{\left(\sin \frac{x}{2}+\cos \frac{x}{2}\right)^{2}}}$

$=\frac{\sin \frac{x}{2}-\cos \frac{x}{2}}{\sin \frac{x}{2}+\cos \frac{x}{2}}$

$=\frac{\tan \frac{x}{2}-1}{\tan \frac{x}{2}+1} \quad$ (dividing by $\cos \frac{x}{2}$ )

$=\frac{\tan \frac{x}{2}-\tan \frac{\pi}{4}}{1+\tan \frac{\pi}{4} \tan \frac{x}{2}}$

$=\tan \left(\frac{x}{2}-\frac{\pi}{4}\right)$

i. e $\sqrt{\frac{1-\sin x}{1+\sin x}}=\tan \left(\frac{x}{2}-\frac{\pi}{4}\right)$