સાબિત કરો કે : $\frac{\sin x+\sin 3 x}{\cos x+\cos 3 x}=\tan 2 x$
It is known that
$\sin A+\sin B=2 \sin \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)$
$\cos A+\cos B=2 \cos \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)$
$\therefore$ $L.H.S.$ $=\frac{\sin x+\sin 3 x}{\cos x+\cos 3 x}$
$=\frac{2 \sin \left(\frac{x+3 x}{2}\right) \cos \left(\frac{x-3 x}{2}\right)}{2 \cos \left(\frac{x+3 x}{2}\right) \cos \left(\frac{x-3 x}{2}\right)}$
$=\frac{\sin 2 x}{\cos 2 x}$
$=\tan 2 x$
$= R . H.S$
$3\,\left[ {{{\sin }^4}\,\left( {\frac{{3\pi }}{2} - \alpha } \right) + {{\sin }^4}\,(3\pi + \alpha )} \right]$ $ - 2\,\left[ {{{\sin }^6}\,\left( {\frac{\pi }{2} + \alpha } \right) + {{\sin }^6}(5\pi - \alpha )} \right] = $
${\sin ^2}\frac{\pi }{8} + {\sin ^2}\frac{{3\pi }}{8} + {\sin ^2}\frac{{5\pi }}{8} + {\sin ^2}\frac{{7\pi }}{8} = $
$\sqrt 2 + \sqrt 3 + \sqrt 4 + \sqrt 6 = . . ..$
જો $\alpha + \beta + \gamma = 2\pi ,$ તો
જો $\theta $ એ લઘુકોણ છે અને $\sin \frac{\theta }{2} = \sqrt {\frac{{x - 1}}{{2x}}} $, તો $\tan \theta = . . .$