${\sin ^2}\frac{\pi }{8} + {\sin ^2}\frac{{3\pi }}{8} + {\sin ^2}\frac{{5\pi }}{8} + {\sin ^2}\frac{{7\pi }}{8} = $

If $\frac{\sqrt{2} \sin \alpha}{\sqrt{1+\cos 2 \alpha}}=\frac{1}{7}$ and $\sqrt{\frac{1-\cos 2 \beta}{2}}=\frac{1}{\sqrt{10}}$ $\alpha, \beta \in\left(0, \frac{\pi}{2}\right),$ then $\tan (\alpha+2 \beta)$ is equal to

$\cos \frac{\pi }{7}\cos \frac{{2\pi }}{7}\cos \frac{{3\pi }}{7} =$

$\tan {3^o} + 2\tan {6^o} + 4\tan {12^o} + 8\cot {24^o} = \cot {\theta ^o}$ then

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

$\cos 20^\circ \cos 40^\circ \cos 80^\circ = $