If 0

(b) $\tan \frac{x}{2}$

We have:

$\frac{y+1}{1-y}=\sqrt{\frac{1+\sin x}{1-\sin x}}$

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

$\Rightarrow \frac{y+1}{1-y}=\sqrt{\frac{\left(\cos \frac{x}{2}+\sin \frac{x}{2}\right)^{2}}{\left(\cos \frac{x}{2}-\sin \frac{x}{2}\right)^{2}}}$

$\Rightarrow \frac{y+1}{1-y}=\frac{\left(\cos \frac{x}{2}+\sin \frac{x}{2}\right)}{\left(\cos \frac{x}{2}-\sin \frac{x}{2}\right)}$ $\left[\because 0

$\Rightarrow \frac{y+1}{1-y}=\frac{\frac{\cos \frac{x}{2}}{\cos \frac{x}{2}}+\frac{\sin \frac{x}{2}}{\cos \frac{x}{2}}}{\frac{\cos \frac{x}{2}}{\cos \frac{x}{2}}-\frac{\sin \frac{x}{2}}{\cos \frac{x}{2}}}$

$\Rightarrow \frac{1+y}{1-y}=\frac{1+\tan \frac{x}{2}}{1-\tan \frac{x}{2}}$

Comparing both the sides:

$y=\tan \frac{x}{2}$