If $0 (a) $\cot \frac{x}{2}$ (b) $\tan \frac{x}{2}$ (c) $\cot \frac{x}{2}+\tan \frac{x}{2}$ (d) $\cot \frac{x}{2}-\tan \frac{x}{2}$
(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: $y=\tan \frac{x}{2}$
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.