निम्नलिखित को सिद्ध कीजिए
$\frac{\sin x+\sin 3 x}{\cos x+\cos 3 x}=\tan 2 x$
It is known that
$\sin A+\sin B=2 \sin \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)$
$\cos A+\cos B=2 \cos \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)$
$\therefore$ $L.H.S.$ $=\frac{\sin x+\sin 3 x}{\cos x+\cos 3 x}$
$=\frac{2 \sin \left(\frac{x+3 x}{2}\right) \cos \left(\frac{x-3 x}{2}\right)}{2 \cos \left(\frac{x+3 x}{2}\right) \cos \left(\frac{x-3 x}{2}\right)}$
$=\frac{\sin 2 x}{\cos 2 x}$
$=\tan 2 x$
$= R . H.S$
$\frac{{\sin 3A - \cos \left( {\frac{\pi }{2} - A} \right)}}{{\cos A + \cos (\pi + 3A)}} = $
$\tan 3A - \tan 2A - \tan A = $
${\sin ^2}\frac{\pi }{8} + {\sin ^2}\frac{{3\pi }}{8} + {\sin ^2}\frac{{5\pi }}{8} + {\sin ^2}\frac{{7\pi }}{8} = $
यदि $\tan A = \frac{1}{2},$ तो $\tan 3A = $
$\cos \left(\frac{2 \pi}{7}\right)+\cos \left(\frac{4 \pi}{7}\right)+\cos \left(\frac{6 \pi}{7}\right)$ का मान बराबर होगा।