Prove the following identity,where the angles involved are acute angles for which the expressions are defined:
$\frac{\sin \theta - 2 \sin^3 \theta}{2 \cos^3 \theta - \cos \theta} = \tan \theta$

(A) Given: $L.H.S. = \frac{\sin \theta - 2 \sin^3 \theta}{2 \cos^3 \theta - \cos \theta}$
Step $1$: Factor out $\sin \theta$ from the numerator and $\cos \theta$ from the denominator.
$L.H.S. = \frac{\sin \theta (1 - 2 \sin^2 \theta)}{\cos \theta (2 \cos^2 \theta - 1)}$
Step $2$: Use the identity $\sin^2 \theta + \cos^2 \theta = 1$,which implies $\cos^2 \theta = 1 - \sin^2 \theta$.
Substitute this into the denominator:
$L.H.S. = \frac{\sin \theta (1 - 2 \sin^2 \theta)}{\cos \theta [2(1 - \sin^2 \theta) - 1]}$
Step $3$: Simplify the expression inside the bracket in the denominator.
$2(1 - \sin^2 \theta) - 1 = 2 - 2 \sin^2 \theta - 1 = 1 - 2 \sin^2 \theta$
Step $4$: Substitute back into the expression.
$L.H.S. = \frac{\sin \theta (1 - 2 \sin^2 \theta)}{\cos \theta (1 - 2 \sin^2 \theta)}$
Step $5$: Cancel the common term $(1 - 2 \sin^2 \theta)$.
$L.H.S. = \frac{\sin \theta}{\cos \theta} = \tan \theta = R.H.S.$
Hence,the identity is proved.