Prove that $\frac{\sin A - 2\sin^3 A}{2\cos^3 A - \cos A} = an A$

Question

Prove that $\frac{\sin A - 2\sin^3 A}{2\cos^3 A - \cos A} = 	an A$

Accepted Answer

Step 1: LHS $= \frac{\sin A - 2\sin^3 A}{2\cos^3 A - \cos A} = \frac{\sin A(1 - 2\sin^2 A)}{\cos A(2\cos^2 A - 1)}$
Step 2: $= \frac{\sin A[1 - 2(1 - \cos^2 A)]}{\cos A[2\cos^2 A - 1]} = \frac{\sin A[2\cos^2 A - 1]}{\cos A[2\cos^2 A - 1]}$
Step 3: $= \frac{\sin A}{\cos A} = 	an A =$ RHS. Hence proved.

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