The value of ${\cos ^2}\,{10^o}\,\, - \,\cos \,\,{10^o}\,\cos \,\,{50^o}\, + \,{\cos ^2}\,{50^o}$ is
$\frac{3}{2}\,(1\, + \,\cos \,{20^o})$
$\frac {3}{4}$
$\frac {3}{2}$
$\frac{3}{4}\,\, + \,\,\cos \,{20^o}$
${\sin ^2}\frac{\pi }{8} + {\sin ^2}\frac{{3\pi }}{8} + {\sin ^2}\frac{{5\pi }}{8} + {\sin ^2}\frac{{7\pi }}{8} = $
Prove that $\sin ^{2} 6 x-\sin ^{2} 4 x=\sin 2 x \sin 10 x$
If $\tan x = \frac{b}{a},$ then $\sqrt {\frac{{a + b}}{{a - b}}} + \sqrt {\frac{{a - b}}{{a + b}}} = $
$2\,{\sin ^2}\beta + 4\,\,\cos \,(\alpha + \beta )\,\,\sin \,\alpha \,\sin \,\beta + \cos \,2\,(\alpha + \beta ) = $
$\sqrt 3 \,{\rm{cosec}}\,{20^o} - \sec \,{20^o} = $