Prove that $\frac{\sin x+\sin 3 x}{\cos x+\cos 3 x}=\tan 2 x$
Prove that $\frac{\sin x+\sin 3 x}{\cos x+\cos 3 x}=\tan 2 x$
It is known that
$\sin A+\sin B=2 \sin \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)$
$\cos A+\cos B=2 \cos \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)$
$\therefore$ $L.H.S.$ $=\frac{\sin x+\sin 3 x}{\cos x+\cos 3 x}$
$=\frac{2 \sin \left(\frac{x+3 x}{2}\right) \cos \left(\frac{x-3 x}{2}\right)}{2 \cos \left(\frac{x+3 x}{2}\right) \cos \left(\frac{x-3 x}{2}\right)}$
$=\frac{\sin 2 x}{\cos 2 x}$
$=\tan 2 x$
$= R . H.S$
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