If $y\left(\alpha \right)=\sqrt{2\left(\frac{\tan \alpha +\cot \alpha }{1+{\tan }^2\alpha }\right)+\frac{1}{{\sin }^2\alpha },}\alpha \in (\frac{3\pi }{4},\pi ),$ then $\frac{\mathrm{dy}}{\mathrm{d\alpha }}\mathrm{at}\mathrm{\alpha }=\frac{5\mathrm{\pi }}{6}$ is 

Given $y=\sqrt{\frac{2(\tan \alpha +\cot \alpha )}{1+{\tan }^2\alpha }+\frac{1}{{\sin }^2\alpha }}$

$=\sqrt{\frac{2({\tan }^2\alpha +1)}{\tan \alpha (1+{\tan }^2\alpha )}+\frac{1}{{\sin }^2\alpha }}=\sqrt{2\cot \alpha +\cos e{c}^2\alpha }$

$=|1+\cot \alpha |=-1-\cot \alpha$ $\therefore \frac{dy}{d\alpha }=\cos e{c}^2\alpha \Rightarrow \frac{dy}{d\alpha }{|}_{\alpha =\frac{5\pi }{6}}=\cos e{c}^2\left(\frac{5\pi }{6}\right)=4$