$$\frac{{\sin \,\theta \, - \,2\,{{\sin }^3}\,\theta }}{{2\,{{\cos }^3}\theta \, - \,\cos \,\theta }}\, = \,\tan \,\theta $$

Solution

LHS $$\dfrac{{\sin \theta  - 2{{\sin }^3}\theta }}{{2{{\cos }^3}\theta  - \cos \theta }}$$

$$ = \dfrac{{\sin \theta \left( {1 - 2{{\sin }^2}\theta } \right)}}{{\cos \theta \left( {2{{\cos }^2}\theta  - 1} \right)}}$$

$$ = \dfrac{{\sin \theta }}{{\cos \theta }}\frac{{}}{{}} = \tan \theta = RHS $$