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.

$3\,\left[ {{{\sin }^4}\,\left( {\frac{{3\pi }}{2} - \alpha } \right) + {{\sin }^4}\,(3\pi + \alpha )} \right]$ $ - 2\,\left[ {{{\sin }^6}\,\left( {\frac{\pi }{2} + \alpha } \right) + {{\sin }^6}(5\pi - \alpha )} \right] = $

કોઈ પણ $\theta \, \in \,\left( {\frac{\pi }{4},\frac{\pi }{2}} \right)$ માટે, $3\,{\left( {\sin \,\theta  - \cos \,\theta } \right)^4} + 6{\left( {\sin \,\theta  + \cos \,\theta } \right)^2} + 4\,{\sin ^6}\,\theta $ =

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

જો $\tan A = \frac{1}{2},\tan B = \frac{1}{3},$ તો $\cos 2A = $

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