if  $y=\sqrt{\left(\frac{1+\cos 2\theta }{1-\cos 2\theta }\right)}\text{, then }\frac{dy}{d\theta }\text{ at }\theta =\frac{3\pi }{4}$ is

$\mathrm{y}=\sqrt{\frac{1+\cos 2\theta }{1-\cos 2\theta }}\Rightarrow \mathrm{y}=\sqrt{\frac{2{\cos }^2\theta }{2{\sin }^2\theta }}=\sqrt{{\cot }^2\theta }$
$\Rightarrow y=\cot \theta$
Differentiate w.r.t $\text{ ' }\theta \text{ ', we get: }\frac{dy}{d\theta }=-{\mathrm{cosec}}^2\theta$

$\begin{array}{r}\text{ Now, }{\left(\frac{\mathrm{dy}}{\mathrm{d}\theta }\right)}_{\theta =\frac{3\pi }{4}}=-{\mathrm{cosec}}^2\left(\frac{3\pi }{4}\right)\\ =-{\mathrm{cosec}}^2\left(\pi -\frac{\pi }{4}\right)=-{\mathrm{cosec}}^2\frac{\pi }{4}=-2\end{array}$