સાબિત કરો કે : $\frac{\cos 4 x+\cos 3 x+\cos 2 x}{\sin 4 x+\sin 3 x+\sin 2 x}=\cot 3 x$
$L.H.S.$ $=\frac{\cos 4 x+\cos 3 x+\cos 2 x}{\sin 4 x+\sin 3 x+\sin 2 x}$
$=\frac{(\cos 4 x+\cos 2 x)+\cos 3 x}{(\sin 4 x+\sin 2 x)+\sin 3 x}$
$=\frac{2 \cos \left(\frac{4 x+2 x}{2}\right) \cos \left(\frac{4 x-2 x}{2}\right)+\cos 3 x}{2 \sin \left(\frac{4 x+2 x}{2}\right) \cos \left(\frac{4 x-2 x}{2}\right)+\sin 3 x}$
$[\because \cos A+\cos B=2 \cos \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right),$
$\sin A+\sin B=2 \sin \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)]$
$=\frac{2 \cos 3 x \cos +\cos 3 x}{2 \sin 3 x \cos x+\sin 3 x}$
$=\frac{\cos 3 x(2 \cos x+1)}{\sin 3 x(2 \cos x+1)}$
$\cot 3 x=R .H .S.$
જો $A + B + C = \pi ,$ તો $\frac{{\cos A}}{{\sin B\sin C}} + \frac{{\cos B}}{{\sin C\sin A}} + \frac{{\cos C}}{{\sin A\sin B}} = $
સાબિત કરો કે : $\frac{\sin x+\sin 3 x}{\cos x+\cos 3 x}=\tan 2 x$
જો $A + B + C = \frac{{3\pi }}{2},$ તો $\cos 2A + \cos 2B + \cos 2C = $
$tan^{-1} (\frac{sin2 -1}{cos2})$ =
$\sin 4\theta $ ને . . . . સ્વરૂપે પણ લખી શકાય.