$\cos \frac{\pi }{5}\cos \frac{{2\pi }}{5}\cos \frac{{4\pi }}{5}\cos \frac{{8\pi }}{5} = $

$\frac{{\tan \,\left( {{\textstyle{{3\,\pi } \over 2}}\,\, - \,\,\alpha } \right)\,\,\,\cos \,\left( {{\textstyle{{3\,\pi } \over 2}}\,\, - \,\,\alpha } \right)}}{{\cos \,(2\,\pi \,\, - \,\alpha )}}$ $+ cos \left( {\alpha \,\, - \,\,\frac{\pi }{2}} \right) \,sin (\pi -\alpha ) + cos (\pi +\alpha ) sin \,\left( {\alpha \,\, - \,\,\frac{\pi }{2}} \right)$ = 

સાબિત કરો કે, $=\frac{\sin 5 x-2 \sin 3 x+\sin x}{\cos 5 x-\cos x}=\tan x$

$\frac{{4\sin {9^o}\sin {{21}^o}\sin {{39}^o}\sin {{51}^o}\sin {{69}^o}\sin {{81}^o}}}{{\sin {{54}^o}}}$ = 

$1 + \cos 2x + \cos 4x + \cos 6x = $

જો $a\tan \theta = b$, તો $a\cos 2\theta + b\sin 2\theta = $