$\tan 7\frac{1}{2}^\circ =...$
$\sqrt 6 + \sqrt 3 + \sqrt 2 - 2$
$\sqrt 6 - \sqrt 3 + \sqrt 2 - 2$
$\sqrt 6 - \sqrt 3 + \sqrt 2 + 2$
$\sqrt 6 - \sqrt 3 - \sqrt 2 - 2$
જો ${\cos ^6}\alpha + {\sin ^6}\alpha + K\,{\sin ^2}2\alpha = 1,$ તો $K =$
જો $\alpha + \beta + \gamma = 2\pi ,$ તો
જો $\frac{\sqrt{2} \sin \alpha}{\sqrt{1+\cos 2 \alpha}}=\frac{1}{7}$ અને $\sqrt{\frac{1-\cos 2 \beta}{2}}=\frac{1}{\sqrt{10}}$ $\alpha, \beta \in\left(0, \frac{\pi}{2}\right),$ તો $\tan (\alpha+2 \beta)$ મેળવો.
જો $\sin \theta + \sin 2\theta + \sin 3\theta = \sin \alpha $અને $\cos \theta + \cos 2\theta + \cos 3\theta = \cos \alpha $, તો $\theta$ મેળવો.
સાબિત કરો કે : $\sin 2 x+2 \sin 4 x+\sin 6 x=4 \cos ^{2} x \sin 4 x$