If $A + B + C = {180^o},$ then the value of $\cot \frac{A}{2} + \cot \frac{B}{2} + \cot \frac{C}{2}$ will be
$2\cot \frac{A}{2}\cot \frac{B}{2}\cot \frac{C}{2}$
$4\cot \frac{A}{2}\cot \frac{B}{2}\cot \frac{C}{2}$
$\cot \frac{A}{2}\cot \frac{B}{2}\cot \frac{C}{2}$
$8\,\cot \frac{A}{2}\cot \frac{B}{2}\cot \frac{C}{2}$
The value of $cot\, x + cot\, (60^o + x) + cot\, (120^o + x)$ is equal to :
If $\tan \theta = t,$ then $\tan 2\theta + \sec 2\theta = $
If $\sin \theta = \frac{1}{2}\left( {\sqrt {\frac{x}{y}\,} + \,\sqrt {\frac{y}{x}} } \right)\,,\,\left( {x,y \in R\, - \{ 0\} } \right)$. Then
If $\tan \alpha = \frac{1}{7}$ and $\sin \beta = \frac{1}{{\sqrt {10} }}\left( {0 < \alpha ,\,\beta < \frac{\pi }{2}} \right)$, then $2\beta $ is equal to
$A, B, C$ are the angles of a triangle, then ${\sin ^2}A + {\sin ^2}B + {\sin ^2}C - 2\cos A\,\cos B\,\cos C = $