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Prove that $\frac{\sin \theta - 2 \sin^3 \theta}{2 \cos^3 \theta - \cos \theta} = \tan \theta$.
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$L.H.S = \frac{\sin\theta-2\sin^3\theta}{2\cos^3\theta-\cos\theta}$
$= \frac{\sin\theta(1-2\sin^2\theta)}{\cos\theta(2\cos^2\theta-1)}$
$= \frac{\tan\theta(1-2\sin^2\theta)}{[2(1-\sin^2\theta)-1]}$
$= \frac{\tan\theta(1-2\sin^2\theta)}{(1-2\sin^2\theta)}$
$= \tan\theta = R.H.S.$
$= \frac{\sin\theta(1-2\sin^2\theta)}{\cos\theta(2\cos^2\theta-1)}$
$= \frac{\tan\theta(1-2\sin^2\theta)}{[2(1-\sin^2\theta)-1]}$
$= \frac{\tan\theta(1-2\sin^2\theta)}{(1-2\sin^2\theta)}$
$= \tan\theta = R.H.S.$