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

The correct answer is:
$\frac{\sqrt{11}+1}{2\sqrt{3}}$

Let us solve the problem step-by-step.
We are given that $\frac{\pi }{2}<x<\pi$, which means $x$ lies in the second quadrant.
We are also given:
$\cot x=\frac{-5}{\sqrt{11}}$

Using the trigonometric identity ${\csc }^{2}x=1+{\cot }^{2}x$, we can find $\csc x$:
${\csc }^{2}x=1+{\left(\frac{-5}{\sqrt{11}}\right)}^2=1+\frac{25}{11}=\frac{36}{11}$
Since $x$ is in the second quadrant, $\sin x$ (and thus $\csc x$) is positive:
$\csc x=\frac{6}{\sqrt{11}}\Rightarrow \sin x=\frac{\sqrt{11}}{6}$

Now we determine $\cos x$:
$\cos x=\cot x·\sin x=(\frac{-5}{\sqrt{11}})·\left(\frac{\sqrt{11}}{6}\right)=-\frac{5}{6}$

Let the expression we want to evaluate be $E$:
$E=\sin \left(\frac{11x}{2}\right)(\sin 6x-\cos 6x)+\cos \left(\frac{11x}{2}\right)(\sin 6x+\cos 6x)$

Expanding the terms:
$E=\sin \left(\frac{11x}{2}\right)\sin 6x-\sin \left(\frac{11x}{2}\right)\cos 6x+\cos \left(\frac{11x}{2}\right)\sin 6x+\cos \left(\frac{11x}{2}\right)\cos 6x$

We can regroup the terms as follows:
$E=[\sin 6x\cos \left(\frac{11x}{2}\right)-\cos 6x\sin \left(\frac{11x}{2}\right)]+[\cos 6x\cos \left(\frac{11x}{2}\right)+\sin 6x\sin \left(\frac{11x}{2}\right)]$

Using the trigonometric identities:
$\sin (A-B)=\sin A\cos B-\cos A\sin B$
and
$\cos (A-B)=\cos A\cos B+\sin A\sin B$
with $A=6x$ and $B=\frac{11x}{2}$, we get:
$E=\sin (6x-\frac{11x}{2})+\cos (6x-\frac{11x}{2})$

Simplifying the angle:
$6x-\frac{11x}{2}=\frac{12x-11x}{2}=\frac{x}{2}$
Thus, the expression simplifies to:
$E=\sin \left(\frac{x}{2}\right)+\cos \left(\frac{x}{2}\right)$

Since $\frac{\pi }{2}<x<\pi$, the half-angle satisfies $\frac{\pi }{4}<\frac{x}{2}<\frac{\pi }{2}$.
This means $\frac{x}{2}$ is in the first quadrant, so both $\sin \left(\frac{x}{2}\right)$ and $\cos \left(\frac{x}{2}\right)$ are positive.
We can use the half-angle formulas:
$\sin \left(\frac{x}{2}\right)=\sqrt{\frac{1-\cos x}{2}}$
and
$\cos \left(\frac{x}{2}\right)=\sqrt{\frac{1+\cos x}{2}}$

Substitute $\cos x=-\frac{5}{6}$:
$\sin \left(\frac{x}{2}\right)=\sqrt{\frac{1-(-\frac{5}{6})}{2}}=\sqrt{\frac{1+\frac{5}{6}}{2}}=\sqrt{\frac{11}{12}}=\frac{\sqrt{11}}{2\sqrt{3}}$
and
$\cos \left(\frac{x}{2}\right)=\sqrt{\frac{1+(-\frac{5}{6})}{2}}=\sqrt{\frac{1-\frac{5}{6}}{2}}=\sqrt{\frac{1}{12}}=\frac{1}{2\sqrt{3}}$

Summing these values, we get: