Consider $f(x)=\tan ^{-1}\left(\sqrt{\frac{1+\sin x}{1-\sin x}}\right)$, $x \in\left(0, \frac{\pi}{2}\right)$. A normal to $y=f(x)$ at $x=\frac{\pi}{6}$ also passes through the point:

$\left(0, \frac{2 \pi}{3}\right)$

We have,

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

$\Rightarrow y =\tan ^{-1}\left(\tan \left(\frac{\pi}{4}+\frac{x}{2}\right)\right) $

$\Rightarrow y =\frac{\pi}{4}+\frac{x}{2}$

$\Rightarrow \frac{d y}{d x}=\frac{1}{2} \Rightarrow \frac{1}{d y / d x}=-2$

When $x =\frac{\pi}{6}, y=\frac{\pi}{4}+\frac{\pi}{12}=\frac{\pi}{3}$

The equation of the normal at $(\pi / 6, \pi / 3)$ is

$y-\frac{\pi}{3}=-2\left(x-\frac{\pi}{6}\right)$ or, $2 x+y-\frac{2 \pi}{3}=0$

Clearly, it passes through $\left(0, \frac{2 \pi}{3}\right)$.