$\left( {\frac{{\sin 2A}}{{1 + \cos 2A}}} \right)\,\left( {\frac{{\cos A}}{{1 + \cos A}}} \right)= $

$\tan \frac{A}{2}$ is equal to

Let $A, B, C$ are three angles such that $sinA + sinB + sinC = 0,$ then

$ \frac {sinAsin BsinC}{(sin 3A+ sin 3B+ sin 3C)}$ (wherever definied) is -

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

$\cos \frac{\pi }{7}\cos \frac{{2\pi }}{7}\cos \frac{{4\pi }}{7} = $

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 $\tan 2\alpha =$