$\sin {20^o}\,\sin {40^o}\,\sin {60^o}\,\sin {80^o} = $
$ - 3/16$
$5/16$
$3/16$
$ - 5/16$
$\frac{{\sec 8A - 1}}{{\sec 4A - 1}} = $
If $A + B + C = {180^o},$ then the value of $(\cot B + \cot C)$ $(\cot C + \cot A)\,\,(\cot A + \cot B)$ will be
$\frac{{\cos A}}{{1 - \sin A}} = $
The expression $[1 - sin (3\pi - \alpha ) + cos (3\pi + \alpha )]$ $\left[ {1\,\, - \,\,\sin \,\left( {\frac{{3\,\pi }}{2}\,\, - \,\,\alpha } \right)\,\, + \,\,\cos \,\left( {\frac{{5\,\pi }}{2}\,\, - \,\,\alpha } \right)} \right]$ when simplified reduces to :
${\sin ^4}\frac{\pi }{8} + {\sin ^4}\frac{{3\pi }}{8} + {\sin ^4}\frac{{5\pi }}{8} + {\sin ^4}\frac{{7\pi }}{8} = $