$\frac{{\sqrt {1 + \sin x} + \sqrt {1 - \sin x} }}{{\sqrt {1 + \sin x} - \sqrt {1 - \sin x} }} = $ (when $x$ lies in $II^{nd}$ quadrant)
$\sin \frac{x}{2}$
$\tan \frac{x}{2}$
$\sec \frac{x}{2}$
${\rm{cosec}}\frac{x}{2}$
In a triangle $ABC,$ the value of $\sin A + \sin B + \sin C$ is
If $\alpha + \beta - \gamma = \pi ,$ then ${\sin ^2}\alpha + {\sin ^2}\beta - {\sin ^2}\gamma = $
$\cos \frac{\pi }{7}\cos \frac{{2\pi }}{7}\cos \frac{{4\pi }}{7} = $
$A, B, C$ are the angles of a triangle, then ${\sin ^2}A + {\sin ^2}B + {\sin ^2}C - 2\cos A\,\cos B\,\cos C = $
If $A + B + C = {180^o},$ then $\frac{{\tan A + \tan B + \tan C}}{{\tan A\,.\,\tan B\,.\,\tan C}} = $