Prove that

$=\frac{\cos 2 x \sin x+\cos 6 x \sin 3 x}{\sin 2 x \sin x+\sin 6 x \sin 3 x}$

$=\frac{2 \cos 2 x \sin x+2 \cos 6 x \sin 3 x}{2 \sin 2 x \sin x+2 \sin 6 x \sin 3 x}$

$=\frac{\sin (2 x+x)-\sin (2 x-x)+\{\sin (6 x+3 x)-\sin (6 x-3 x)\}}{\cos (2 x-x)-\cos (2 x+x)+\cos (6 x-3 x)-\cos (6 x+3 x)}$

$=\frac{\sin 3 x-\sin x+\sin 9 x-\sin 3 x}{\cos x-\cos 3 x+\cos 3 x-\cos 9 x}$

$=\frac{\sin 9 x-\sin x}{\cos x-\cos 9 x}$

$=\frac{2 \cos \frac{9 x+x}{2} \sin \frac{9 x-x}{2}}{-2 \sin \frac{x+9 x}{2} \sin \frac{x-9 x}{2}}$

$=\frac{2 \cos \frac{9 x+x}{2} \sin \frac{9 x-x}{2}}{2 \sin \frac{x+9 x}{2} \sin \frac{9 x^{2}-x}{2}}$

$=\frac{\cos 5 x \sin 4 x}{\sin 5 x \cos 4 x}$

$=\cot 5 x$

Using the formulas,

$2 \cos A \sin B=\sin (A+B)-\sin (A-B)$

$2 \sin A \sin B=\cos (A-B)-\cos (A+B)$