The expression $\frac{{{{\tan }^2}20^\circ  - {{\sin }^2}20^\circ }}{{{{\tan }^2}20^\circ \,\cdot\,{{\sin }^2}20^\circ }}$ simplifies to

The value of $\frac{1}{4} \,\,tan \frac{\pi}{8} +\frac{1}{8} \,\,tan \frac{\pi}{16}+\frac{1}{16} \,\,tan \frac{\pi}{32}+.\,.\,.\,\infty  $ terms is equal to-

$\left( {\frac{{\sin 2A}}{{1 + \cos 2A}}} \right)\,\left( {\frac{{\cos A}}{{1 + \cos A}}} \right)= $

If $\alpha + \beta - \gamma = \pi ,$ then ${\sin ^2}\alpha + {\sin ^2}\beta - {\sin ^2}\gamma = $

Show that

$\tan 3 x \tan 2 x \tan x=\tan 3 x-\tan 2 x-\tan x$

If $\sin \theta + \sin 2\theta + \sin 3\theta = \sin \alpha $and $\cos \theta + \cos 2\theta + \cos 3\theta = \cos \alpha $, then $\theta$ is equal to