Prove that: $ \frac{\sin A - 2 \sin^3 A} {2 \cos^3 A – \cos A} = \tan A $

Prove that: $ \frac{\sin A - 2 \sin^3 A} {2 \cos^3 A – \cos A} = \tan A$

$$ LHS = \frac \sin A - 2 sin^3A} {2cos^3A - cos A} $$

$$ = \frac{\sin A(1 - 2 sin ^2A)}{cos A (2 cos ^2A -1 )} $$

$$ = \frac{\sin A(1-2(1 - \cos^2A))}{ \cos A(2 cos^2 A - 1)} $$

$$ = \tan A\frac{(2cos^2 A - 1)}{(2cos^2 A -1)} $$

$$ = \tan A= RHS $$