If $\alpha + \beta - \gamma = \pi ,$ then ${\sin ^2}\alpha + {\sin ^2}\beta - {\sin ^2}\gamma = $

The expression,$\frac{{\tan \,\left( {{\textstyle{{3\,\pi } \over 2}}\,\, - \,\,\alpha } \right)\,\,\,\cos \,\left( {{\textstyle{{3\,\pi } \over 2}}\,\, - \,\,\alpha } \right)}}{{\cos \,(2\,\pi \,\, - \,\alpha )}}$ $+ cos \left( {\alpha \,\, - \,\,\frac{\pi }{2}} \right) \,sin (\pi -\alpha ) + cos (\pi +\alpha ) sin \,\left( {\alpha \,\, - \,\,\frac{\pi }{2}} \right)$ when simplified reduces to :

$(\sec 2A + 1){\sec ^2}A = $

If $\alpha$, $\beta$,$\gamma$ are positive number such that $\alpha + \beta = \pi$  and $\beta  + \gamma = \alpha$, then $tan\ \alpha$ is equal to - (where $\gamma  \ne n\pi ,n \in I$ )

The value of $\sin \frac{\pi }{{14}}\sin \frac{{3\pi }}{{14}}\sin \frac{{5\pi }}{{14}}\sin \frac{{7\pi }}{{14}}\sin \frac{{9\pi }}{{14}}\sin \frac{{11\pi }}{{14}}\sin \frac{{13\pi }}{{14}}$ is equal to

If $a\,\cos 2\theta + b\,\sin 2\theta = c$  has $\alpha$ and $\beta$ as its solution, then the value of $\tan \alpha + \tan \beta $ is