Show that

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

If $\cos x + \cos y + \cos \alpha = 0$ and $\sin x + \sin y + \sin \alpha = 0,$ then $\cot \,\left( {\frac{{x + y}}{2}} \right) = $

If $\frac{{\cos x}}{a} = \frac{{\cos (x + \theta )}}{b} = \frac{{\cos (x + 2\theta )}}{c} = \frac{{\cos (x + 3\theta )}}{d} \, ,$ then $\left( {\frac{{a + c}}{{b + d}}} \right)$ is equal to :-

In any triangle $ABC ,$ ${\sin ^2}\frac{A}{2} + {\sin ^2}\frac{B}{2} + {\sin ^2}\frac{C}{2}$ is equal to

The expression,$\frac{{\tan \,\left( {{\textstyle{{3\,\pi } \over 2}}\,\, - \,\,\alpha } \right)\,\,\,\cos \,\left( {{\textstyle{{3\,\pi } \over 2}}\,\, - \,\,\alpha } \right)}}{{\cos \,(2\,\pi \,\, - \,\alpha )}}$ $+ cos \left( {\alpha \,\, - \,\,\frac{\pi }{2}} \right) \,sin (\pi -\alpha ) + cos (\pi +\alpha ) sin \,\left( {\alpha \,\, - \,\,\frac{\pi }{2}} \right)$ when simplified reduces to :

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