Prove the identity: $\;$ $\dfrac{\sin \theta - 2 \sin^3 \theta}{2 \cos^3 \theta - \cos \theta} = \tan \theta$
$\begin{aligned}
LHS & = \dfrac{\sin \theta - 2 \sin^3 \theta}{2 \cos^3 \theta - \cos \theta} \\\\
& = \dfrac{\sin \theta \left(1 - 2 \sin^2 \theta\right)}{\cos \theta \left(2 \cos^2 \theta - 1\right)} \\\\
& = \dfrac{\sin \theta \left(\sin^2 \theta + \cos^2 \theta - 2 \sin^2 \theta\right)}{\cos \theta \left(2 \cos^2 \theta - \sin^2 \theta - \cos^2 \theta\right)} \;\;\; \left[\text{Note: } \sin^2 \theta + \cos^2 \theta = 1\right] \\\\
& = \dfrac{\sin \theta \left(\cos^2 \theta - \sin^2 \theta\right)}{\cos \theta \left(\cos^2 \theta - \sin^2 \theta\right)} \\\\
& = \dfrac{\sin \theta}{\cos \theta} \\\\
& = \tan \theta = RHS
\end{aligned}$
Hence proved.