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

Let’s start from LHS LHS = $\frac{\sin A - 2\sin^3A}{2\cos^3A-\cos A}$ = $\frac{\sin A (1- 2\sin^2 A)}{\cos A(2\cos^2A-1)}$ = $(\frac{\sin A}{\cos A})(\frac{1 - 2\sin^2A}{2\cos^2A-1})$ We know that sin2A + cos2A = 1 $\therefore$ LHS = $(\tan A)(\frac{(\sin^2A + \cos^2A) - 2\sin^2A}{2\cos^2A -(\sin^2A + \cos^2A)})$ = $(\tan A)(\frac{\cos^2A - \sin^2A}{\cos^2A - \sin^2A})$ =tan A ........ RHS Hence proved!