Answer:
Option B
Explanation:
We have
$f(x)= tan^{-1}\sqrt{\frac{1+\sin x}{1-\sin x}},x\epsilon(0,\frac{\pi}{2})$
$f(x)= tan^{-1}\sqrt{\frac{(\cos\frac{x}{2}+\sin\frac{x}{2})^{2}}{(\cos\frac{x}{2}-\sin \frac{x}{2})^{2}}}$
$f(x)= tan^{-1}({\frac{\cos\frac{x}{2}+\sin\frac{x}{2}^{}}{\cos\frac{x}{2}-\sin \frac{x}{2}^{}}})$
( $\cos \frac{x}{2}>\sin \frac{x}{2} for$ $0< \frac{x}{2}. \frac{\pi}{4} $
$=\tan^{-1}(\frac{1+\tan\frac{x}{2}}{1-\tan\frac{x}{2}})$
$=\tan^{-1}[\tan(\frac{\pi}{4}+\frac{x}{2})]=\frac{\pi}{4}+\frac{x}{2}$
$\Rightarrow f'(x)=\frac{1}{2}\Rightarrow f'(\frac{\pi}{6})=\frac{1}{2}$
Now, equation of normal at x= $\frac{\pi}{6}$ is given by
$(y-f(\frac{\pi}{6}))=-2(x-\frac{\pi}{6})$
$\Rightarrow (y-\frac{\pi}{3})=-2(x-\frac{\pi}{6})$
$[\because f(\frac{\pi}{6})=\frac{\pi}{4}+\frac{\pi}{12}=\frac{4\pi}{12}=\frac{\pi}{3}$
which passes through (0, $\frac{2\pi}{3}$)
