Prove that: $\frac{\sin 5 x+\sin 3 x}{\cos 5 x+\cos 3 x}=\tan 4 x$
Prove that: $\frac{\sin 5 x+\sin 3 x}{\cos 5 x+\cos 3 x}=\tan 4 x$
It is known that
$\sin A+\sin B=2 \sin \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)$
$\cos A+\cos B=2 \cos \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)$
$\therefore$ $L.H.S.$ $=\frac{\sin 5 x+\sin 3 x}{\cos 5 x+\cos 3 x}$
$=\frac{2 \sin \left(\frac{5 x+3 x}{2}\right) \cdot \cos \left(\frac{5 x-3 x}{2}\right)}{2 \cos \left(\frac{5 x+3 x}{2}\right) \cdot \cos \left(\frac{5 x-3 x}{2}\right)}$
$=\frac{2 \sin 4 x \cdot \cos x}{2 \cos 4 x \cdot \cos x}$
$=\tan 4 x=R . H . S$
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- [IIT 1985]
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