Mathematics — Standard (041) — CBSE Class 10 board question

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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• Key step: Factor $\sin A$ from numerator and $\cos A$ from denominator first.
• Core identity used: $\sin^2 A + \cos^2 A = 1$, to show both brackets simplify to the same expression $(\cos^2 A - \sin^2 A)$, which cancels.
• Examiners award 1 mark for correct factoring, 1 mark for applying the identity to both brackets, and 1 mark for the final cancellation and conclusion. Don't skip showing that both expressions equal $\cos^2A - \sin^2A$.