Answer
$\dfrac{\sin 3 x+\sin 3x+\sin 5x}{\cos x+\cos 3x +\cos 5x}=\tan 3x$
Work Step by Step
Consider LHS: $\dfrac{\sin 3 x+\sin 3x+\sin 5x}{\cos x+\cos 3x +\cos 5x}=\dfrac{(\sin x+\sin 5x)+\sin 3x}{(\cos x+\cos 5x) +\cos 3x}$
or, $\dfrac{(\sin x+\sin 5x)+\sin 3x}{(\cos x+\cos 5x) +\cos 3x}=\dfrac{2 \sin 3x \cos 2x +\sin 3x}{2 \cos 3x \cos 2x +\cos 3x}$
or, $\dfrac{2 \sin 3x \cos 2x +\sin 3x}{2 \cos 3x \cos 2x +\cos 3x}=\dfrac{\sin 3x}{ \cos 3x}$
Thus, $\dfrac{\sin 3 x+\sin 3x+\sin 5x}{\cos x+\cos 3x +\cos 5x}=\tan 3x$
Hence, the result has been proved.
