સાબિત કરો કે : $\frac{\sin x+\sin 3 x}{\cos x+\cos 3 x}=\tan 2 x$
It is known that
$\sin A+\sin B=2 \sin \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)$
$\cos A+\cos B=2 \cos \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)$
$\therefore$ $L.H.S.$ $=\frac{\sin x+\sin 3 x}{\cos x+\cos 3 x}$
$=\frac{2 \sin \left(\frac{x+3 x}{2}\right) \cos \left(\frac{x-3 x}{2}\right)}{2 \cos \left(\frac{x+3 x}{2}\right) \cos \left(\frac{x-3 x}{2}\right)}$
$=\frac{\sin 2 x}{\cos 2 x}$
$=\tan 2 x$
$= R . H.S$
જો $\cos \theta = \frac{1}{2}\left( {a + \frac{1}{a}} \right),$ તો $\cos 3\theta = . . .$
જો $\sin A = n\sin B,$ તો $\frac{{n - 1}}{{n + 1}}\tan \,\frac{{A + B}}{2} = $
$2 \sin \left(\frac{\pi}{22}\right) \sin \left(\frac{3 \pi}{22}\right) \sin \left(\frac{5 \pi}{22}\right) \sin \left(\frac{7 \pi}{22}\right) \sin \left(\frac{9 \pi}{22}\right)$ =
${\cos ^2}A{(3 - 4{\cos ^2}A)^2} + {\sin ^2}A{(3 - 4{\sin ^2}A)^2} = $
$\sqrt 2 + \sqrt 3 + \sqrt 4 + \sqrt 6 = . . ..$