If $A$ lies in the third quadrant and $3\,\tan A - 4 = 0,$ then $5\,\sin 2A + 3\,\sin A + 4\,\cos A = $

Given that $\cos \left( {\frac{{\alpha - \beta }}{2}} \right) = 2\cos \left( {\frac{{\alpha + B}}{2}} \right)$, then $\tan \frac{\alpha }{2}\tan \frac{\beta }{2} $ is equal to

If $sin t + cos t = \frac{1}{5}$ then $tan \frac{t}{2}$ is equal to :

$\frac{{\sec 8A - 1}}{{\sec 4A - 1}} = $

$\cos \frac{{2\pi }}{{15}}\cos \frac{{4\pi }}{{15}}\cos \frac{{8\pi }}{{15}}\cos \frac{{16\pi }}{{15}}  =$

If $A + B + C = {180^o},$ then $\frac{{\tan A + \tan B + \tan C}}{{\tan A\,.\,\tan B\,.\,\tan C}} = $