$x \in\left(\frac{\pi}{2}, \pi\right)$

$\left(\sin \frac{11 x}{2}\right)(\sin 6 x-\cos 6 x)+\left(\cos \frac{11 x}{2}\right)(\sin 6 x+\cos 6 x)$

$=\left\{\sin 6 x \sin \frac{11 x}{2}+\cos \frac{11 x}{2} \cos 6 x\right\}$

$=\cos \left(6 x-\frac{11 x}{2}\right)+\sin \left(6 x-\frac{11 x}{2}\right)$

$=\cos \frac{x}{2}+\sin \frac{x}{2}$

$=\frac{1}{2 \sqrt{3}}+\frac{\sqrt{11}}{2 \sqrt{3}}$

$=\frac{\sqrt{11}+1}{2 \sqrt{3}} \Rightarrow \text { Option (B) is correct. }$

$\cot x=-\frac{5}{\sqrt{11}}$

$\frac{1-\tan ^2 \frac{x}{2}}{2 \tan \frac{x}{2}}=-\frac{5}{\sqrt{11}}$

$\tan \frac{x}{2}=\sqrt{11},-\frac{1}{\sqrt{11}}$

$\tan \frac{x}{2}=\sqrt{11}$, As $\frac{\pi}{4}<\frac{x}{2}<\frac{\pi}{2}$