$\tan 3A - \tan 2A - \tan A = $

 $tan^{-1} (\frac{sin2 -1}{cos2})$ = 

સાબિત કરો કે : $\cot 4 x(\sin 5 x+\sin 3 x)=\cot x(\sin 5 x-\sin 3 x)$

${\sin ^4}\frac{\pi }{4} + {\sin ^4}\frac{{3\pi }}{8} + {\sin ^4}\frac{{5\pi }}{8} + {\sin ^4}\frac{{7\pi }}{8} = $

જો $\cos x + \cos y + \cos \alpha = 0$ અને $\sin x + \sin y + \sin \alpha = 0,$ તો $\cot \,\left( {\frac{{x + y}}{2}} \right) = $

$\cos 2(\theta + \phi ) - 4\cos (\theta + \phi )\sin \theta \sin \phi + 2{\sin ^2}\phi = $