Let \(\frac{\pi}{2} < x < \pi\) be such that \(\cot x=\frac{-5}{\sqrt{11}}\). Then \(\left(\sin \frac{11 x}{2}\right)(\sin 6 x-\cos 6 x)+\left(\cos \frac{11 x}{2}\right)(\sin 6 x+\cos 6 x)\) is equal to

Given,
\(\begin{aligned} & \cot x=-\frac{5}{\sqrt{11}} \\
& \frac{1-\tan ^2 \frac{x}{2}}{2 \tan \frac{x}{2}}=-\frac{5}{\sqrt{11}} \\
& \tan \frac{x}{2}=\sqrt{11},-\frac{1}{\sqrt{11}} \\
& \text{So, } \tan \frac{x}{2}=\sqrt{11}, \text { As } \frac{\pi}{4} < \frac{x}{2} < \frac{\pi}{2}\end{aligned}\)
\(\begin{aligned} & \text{As, } x \in\left(\frac{\pi}{2}, \pi\right) \\
& \text{Now, } \left(\sin \frac{11 x}{2}\right)(\sin 6 x-\cos 6 x)+\left(\cos \frac{11 x}{2}\right)(\sin 6 x+\cos 6 x) \\
& =\left\{\sin 6 x \sin \frac{11 x}{2}+\cos \frac{11 x}{2} \cos 6 x \right\} \\
& =\cos \left(6 x-\frac{11 x}{2}\right)+\sin \left(6 x-\frac{11 x}{2}\right) \\
& =\cos \frac{x}{2}+\sin \frac{x}{2} \\
& =\frac{1}{2 \sqrt{3}}+\frac{\sqrt{11}}{2 \sqrt{3}} \\
& =\frac{\sqrt{11}+1}{2 \sqrt{3}} \Rightarrow \text { Option (2) is correct. }\end{aligned}\)