Prove the identity: $\dfrac{\sin A - 2 \sin^3 A}{2 \cos^3 A - \cos A} = \tan A$.
Step-by-Step Solution
Key Concept: Factor out $\sin A$ in numerator and $\cos A$ in denominator, use $1 - 2\sin^2 A = 2\cos^2 A - 1$.
\text{LHS} = \dfrac{\sin A (1 - 2 \sin^2 A)}{\cos A (2 \cos^2 A - 1)}. [1.0 Mark]
Since $1 - 2 \sin^2 A = (\sin^2 A + \cos^2 A) - 2 \sin^2 A = \cos^2 A - \sin^2 A$, and $2 \cos^2 A - 1 = 2 \cos^2 A - (\sin^2 A + \cos^2 A) = \cos^2 A - \sin^2 A$. [1.0 Mark]
Numerator bracket and denominator bracket are identical and cancel out: $\dfrac{\sin A}{\cos A} = \tan A = \text{RHS}$. Proved! [1.0 Mark]
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🎯 Official CBSE Marking Scheme:
Factoring out $\sin A$ and $\cos A$: 1.0 Mark
Showing $1 - 2\sin^2 A = 2\cos^2 A - 1 = \cos^2 A - \sin^2 A$: 1.0 Mark
Cancelling brackets to get $\tan A$: 1.0 Mark
Correct Answer: