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Question:

If $0<x<\frac{\pi}{2}$, and if $\frac{y+1}{1-y}=\sqrt{\frac{1+\sin x}{1-\sin x}}$, then $y$ is equal to

(a) $\cot \frac{x}{2}$

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

(c) $\cot \frac{x}{2}+\tan \frac{x}{2}$

(d) $\cot \frac{x}{2}-\tan \frac{x}{2}$

Solution:

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

We have:

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

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

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

$\Rightarrow \frac{y+1}{1-y}=\frac{\left(\cos \frac{x}{2}+\sin \frac{x}{2}\right)}{\left(\cos \frac{x}{2}-\sin \frac{x}{2}\right)}$ $\left[\because 0<x<\frac{\pi}{2} \Rightarrow 0<\frac{x}{2}<\frac{\pi}{4}, 0\right.$ to $\frac{\pi}{4} \cos x$ is greater than $\left.\sin x\right]$

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

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

Comparing both the sides:

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