Prove that $=\frac{\sin 5 x-2 \sin 3 x+\sin x}{\cos 5 x-\cos x}=\tan x$

Prove that $\cos ^{2} 2 x-\cos ^{2} 6 x=\sin 4 x \sin 8 x$

$3\,\left[ {{{\sin }^4}\,\left( {\frac{{3\pi }}{2} - \alpha } \right) + {{\sin }^4}\,(3\pi + \alpha )} \right]$ $ - 2\,\left[ {{{\sin }^6}\,\left( {\frac{\pi }{2} + \alpha } \right) + {{\sin }^6}(5\pi - \alpha )} \right] = $

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

If $3\cos \theta  + 4\sin \theta  = 5$ then $3\sin \theta  - 4\cos \theta $ is

If $\sin A + \cos A = \sqrt 2 ,$ then ${\cos ^2}A = $