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Prove that $\frac{\sin A - 2 \sin^3 A}{2 \cos^3 A - \cos A} = \tan A$
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Sol. LHS = $\frac{\sin A - 2 \sin^3 A}{2 \cos^3 A - \cos A} = \frac{\sin A (1 – 2 \sin^2 A)}{\cos A (2 \cos^2 A – 1)}$
$= \frac{\sin A[1 - 2(1 – \cos^2 A)]}{\cos A [2 \cos^2 A- - 1]} = \frac{\sin A[1 – 2 + 2 \cos^2 A]}{\cos A[2 \cos^2 A - 1]}$
$= \frac{\sin A[2 \cos^2 A - 1]}{\cos A [2 \cos^2 A – 1]} = \tan A = RHS$
$= \frac{\sin A[1 - 2(1 – \cos^2 A)]}{\cos A [2 \cos^2 A- - 1]} = \frac{\sin A[1 – 2 + 2 \cos^2 A]}{\cos A[2 \cos^2 A - 1]}$
$= \frac{\sin A[2 \cos^2 A - 1]}{\cos A [2 \cos^2 A – 1]} = \tan A = RHS$