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

LHS $= \dfrac{\sin\theta - 2\sin^3\theta}{2\cos^3\theta - \cos\theta}$

Take $\sin\theta$ common from numerator and $\cos\theta$ common from denominator:

$$= \frac{\sin\theta(1 - 2\sin^2\theta)}{\cos\theta(2\cos^2\theta - 1)}$$

Since $\sin^2\theta + \cos^2\theta = 1$, we have $\sin^2\theta = 1 - \cos^2\theta$, so:

$$1 - 2\sin^2\theta = 1 - 2(1-\cos^2\theta) = 2\cos^2\theta - 1$$

Substituting:

$$= \frac{\sin\theta\,(2\cos^2\theta - 1)}{\cos\theta\,(2\cos^2\theta - 1)} = \frac{\sin\theta}{\cos\theta} = \tan\theta = \textbf{ RHS}$$

Hence proved.

Source: Exercise 8.3, Q.4(vii), Chapter 8 — Introduction to Trigonometry

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• The key step is factoring $\sin\theta$ from the numerator and $\cos\theta$ from the denominator, leaving $(1-2\sin^2\theta)$ and $(2\cos^2\theta-1)$.
• Then use the identity $\sin^2\theta = 1-\cos^2\theta$ to show both brackets are equal, so they cancel.
• Examiners award 1 mark for correct factoring, 1 mark for applying the identity to show the brackets match, and 1 mark for the final cancellation giving $\tan\theta$.