The value of $tan^{-1} (\frac{sin2 -1}{cos2})$ is equal to:-
$\frac{\pi}{2} - 1$
$2 - \frac{\pi}{2}$
$1- \frac{\pi}{4}$
$ \frac{\pi}{4}-1$
$\tan \alpha + 2\tan 2\alpha + 4\tan 4\alpha + 8\cot \,8\alpha = $
${\sin ^4}\frac{\pi }{8} + {\sin ^4}\frac{{3\pi }}{8} + {\sin ^4}\frac{{5\pi }}{8} + {\sin ^4}\frac{{7\pi }}{8} = $
$(\sec 2A + 1){\sec ^2}A = $
The expression $\frac{{\cos 6x + 6\cos 4x + 15\cos 2x + 10}}{{\cos 5x + 5\cos 3x + 10\cos x}}$ is equal to
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 $\tan 2\alpha =$