LHS $= \dfrac{\sin A - 2\sin^3 A}{2\cos^3 A - \cos A}$
Take $\sin A$ common from numerator and $\cos A$ common from denominator:
$$= \frac{\sin A(1 - 2\sin^2 A)}{\cos A(2\cos^2 A - 1)}$$
Since $\sin^2 A + \cos^2 A = 1$, we have $1 - \sin^2 A = \cos^2 A$, so:
$$1 - 2\sin^2 A = \sin^2 A + \cos^2 A - 2\sin^2 A = \cos^2 A - \sin^2 A$$
$$2\cos^2 A - 1 = 2\cos^2 A - (\sin^2 A + \cos^2 A) = \cos^2 A - \sin^2 A$$
Therefore:
$$\text{LHS} = \frac{\sin A(\cos^2 A - \sin^2 A)}{\cos A(\cos^2 A - \sin^2 A)} = \frac{\sin A}{\cos A} = \tan A = \text{RHS}$$
Hence proved.
Source: Exercise 8.3, Q4(vii); Section 8.4 Trigonometric Identities, Chapter 8
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