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 $\tan A = \frac{{1 - \cos B}}{{\sin B}},$ હોય તો $\tan 2A$ અને $\tan B$ નો સંબંધ મેળવો..

$cos\, \frac{\pi }{{10}} \,cos\, \frac{2\pi }{{10}} \,cos\,\frac{4\pi }{{10}}\, cos\,\frac{8\pi }{{10}}\, cos\,\frac{16\pi }{{10}}$ = 

$\sqrt 2 + \sqrt 3 + \sqrt 4 + \sqrt 6 = . . ..$

જો $\cos \theta = \frac{1}{2}\left( {a + \frac{1}{a}} \right),$ તો $\cos 3\theta  = . . .$

 $cot\, x + cot\, (60^o  + x) + cot\, (120^o  + x)$  =