We have, $y(x)=\cot ^{-1}\left(\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right)$ $=\cot ^{-1} \frac{\left|\cos \frac{x}{2}+\sin \frac{x}{2}\right|+\left|\cos \frac{x}{2}-\sin \frac{x}{2}\right|}{\left|\cos \frac{x}{2}+\sin \frac{x}{2}\right|-\left|\cos \frac{x}{2}-\sin \frac{x}{2}\right|}$ $\left[\text { as } \cos ^2 \frac{x}{2}+\sin ^2 \frac{x}{2}=1\right.$ and $\left.\sin x=2 \sin \frac{x}{2} \cos \frac{x}{2}\right]$ $\begin{aligned} & =\cot ^{-1}\left(\frac{\cos \frac{x}{2}+\sin \frac{x}{2}+\sin \frac{x}{2}-\cos \frac{x}{2}}{\cos \frac{x}{2}+\sin \frac{x}{2}-\sin \frac{x}{2}+\cos \frac{x}{2}}\right) \forall x \in\left(\frac{\pi}{2}, \pi\right) \\ & \end{aligned}$ $=\cot ^{-1}\left(\frac{\sin \frac{x}{2}}{\cos \frac{x}{2}}\right)=\cot ^{-1}\left(\tan \frac{x}{2}\right)$ $=\frac{\pi}{2}-\tan ^{-1}\left(\tan \frac{x}{2}\right)$ $\begin{aligned} & \therefore y^{\prime}(x)=\frac{\pi}{2}-\frac{x}{2} \\\\ & \Rightarrow \frac{d y}{d x}=y^{\prime}(x)=-\frac{1}{2}=y^{\prime}\left(\frac{5 \pi}{6}\right) \end{aligned}$