Let $\frac{\pi}{2} < x < \pi$ be such that $\cot x=\frac{-5}{\sqrt{11}}$. Then $\left(\sin \frac{11 x}{2}\right)(\sin 6 x-\cos 6 x)+\left(\cos \frac{11 x}{2}\right)(\sin 6 x+\cos 6 x)$ is equal to

$\frac{{\cos A}}{{1 - \sin A}} = $

If $\cos \left( {\alpha + \beta } \right) = \frac{4}{5}$ and $\sin \left( {\alpha - \beta } \right) = \frac{5}{{13}}$,where $0 \le \alpha ,\beta \le \frac{\pi }{4}$ . Then $\tan 2\alpha =$ 

If $\tan \,(A + B) = p,\,\,\tan \,(A - B) = q,$ then the value of $\tan \,2A$ in terms of $p$ and $q$ is

If $A + B + C = \frac{{3\pi }}{2},$ then $\cos 2A + \cos 2B + \cos 2C = $

If $\sin \theta  = \frac{1}{2}\left( {\sqrt {\frac{x}{y}\,}  + \,\sqrt {\frac{y}{x}} } \right)\,,\,\left( {x,y \in R\, - \{ 0\} } \right)$. Then