The value of ${\cos ^2}\,{10^o}\,\, - \,\cos \,\,{10^o}\,\cos \,\,{50^o}\, + \,{\cos ^2}\,{50^o}$ is
$\frac{3}{2}\,(1\, + \,\cos \,{20^o})$
$\frac {3}{4}$
$\frac {3}{2}$
$\frac{3}{4}\,\, + \,\,\cos \,{20^o}$
If $a\,\cos 2\theta + b\,\sin 2\theta = c$ has $\alpha$ and $\beta$ as its solution, then the value of $\tan \alpha + \tan \beta $ is
If $A + B + C = {270^o},$ then $\cos \,2A + \cos 2B + \cos 2C + 4\sin A\,\sin B\,\sin C = $
Prove that $=\frac{\sin 5 x-2 \sin 3 x+\sin x}{\cos 5 x-\cos x}=\tan x$
$\frac{1}{{\tan 3A - \tan A}} - \frac{1}{{\cot 3A - \cot A}} = $
$\frac{{\tan \,\,\left( {x\,\, - \,\,{\textstyle{\pi \over 2}}} \right)\,\,.\,\,\cos \,\,\left( {{\textstyle{{3\pi } \over 2}}\,\, + \,\,x} \right)\,\, - \,\,{{\sin }^3}\,\left( {{\textstyle{{7\pi } \over 2}}\,\, - \,\,x} \right)}}{{\cos \,\,\left( {x\,\, - \,\,{\textstyle{\pi \over 2}}} \right)\,\,.\,\,\tan \,\,\left( {{\textstyle{{3\pi } \over 2}}\,\, + \,\,x} \right)}}$ when simplified reduces to :