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

Imagine you are standing at the edge of a vast, complex landscape. You are presented with an expression that looks like a tangled mess of trigonometric functions:

In the world of JEE Advanced, complexity is often just a mask for elegance. Let us peel back that mask together.

Before we touch the algebra, we must understand our environment. We are given the constraint $\frac{\pi }{2}<x<\pi$.

This tells us that $x$ is in the Second Quadrant. In this territory, $\sin x$ is positive, but $\cos x$ is negative. We are also given $\cot x=-\frac{5}{\sqrt{11}}$.

By constructing a right-angled triangle where the base is $-5$ and the perpendicular is $\sqrt{11}$, we find the hypotenuse using the Pythagorean theorem:

Thus, our anchor value is $\cos x=-\frac{5}{6}$.

Direct substitution is a path to madness. Instead, we expand the expression $E$:

Now, we regroup the terms to identify the standard trigonometric identities $\sin (A-B)=\sin A\cos B-\cos A\sin B$ and $\cos (A-B)=\cos A\cos B+\sin A\sin B$.

Rearranging the terms yields:

The magic happens instantly. The first bracket simplifies to $\sin (6x-\frac{11x}{2})$ and the second to $\cos (6x-\frac{11x}{2})$. This reduces the entire expression to:

We have turned a mountain into a molehill. Since $\frac{\pi }{2}<x<\pi$, it follows that $\frac{\pi }{4}<\frac{x}{2}<\frac{\pi }{2}$, placing us in the First Quadrant where both sine and cosine are positive.

Using the half-angle formulas:

Similarly, for the cosine component:

Adding these together, we arrive at the final result: