Prove that $\displaystyle \frac{\sin \theta-2 \sin ^{3} \theta}{2 \cos ^{3} \theta-\cos \theta}=\tan \theta$.

\[\begin{aligned} \text { L.H.S }= & \frac{\sin \theta-2 \sin ^{3} \theta}{2 \cos ^{3} \theta-\cos \theta} \\ = & \frac{\sin \theta\left(1-2 \sin ^{2} \theta\right)}{\cos \theta\left(2 \cos ^{2} \theta-1\right)} \\ = & \frac{\tan \theta\left(1-2 \sin ^{2} \theta\right)}{\left[2\left(1-\sin ^{2} \theta\right)-1\right]} \\ = & \frac{\tan \theta\left(1-2 \sin ^{2} \theta\right)}{\left(1-2 \sin ^{2} \theta\right)} \\ & =\tan \theta=\text { R.H.S. } \end{aligned} \]