Mathematics — CBSE Class 10 board question

LHS $= \dfrac{\cos\theta - 2\cos^3\theta}{2\sin^3\theta - \sin\theta} \cdot (-1) + \cot\theta$

Wait — take LHS directly:

$$\text{LHS} = \frac{\cos\theta - 2\cos^3\theta}{\sin\theta - 2\sin^3\theta} + \cot\theta$$

Step 1: Factor numerator and denominator.

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

Step 2: Combine over common denominator $\sin\theta(1 - 2\sin^2\theta)$:

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

$$= \frac{\cos\theta\bigl[(1 - 2\cos^2\theta) + (1 - 2\sin^2\theta)\bigr]}{\sin\theta(1 - 2\sin^2\theta)}$$

Step 3: Simplify the bracket using $\sin^2\theta + \cos^2\theta = 1$:

$$= \frac{\cos\theta\bigl[2 - 2(\sin^2\theta + \cos^2\theta)\bigr]}{\sin\theta(1 - 2\sin^2\theta)} = \frac{\cos\theta\,[2 - 2(1)]}{\sin\theta(1 - 2\sin^2\theta)} = \frac{\cos\theta \cdot 0}{\sin\theta(1-2\sin^2\theta)} = 0 = \textbf{RHS}$$

Hence proved. $\blacksquare$

Source: Chapter 8, Section 8.4 Trigonometric Identities

---

• Key identity used: $\sin^2\theta + \cos^2\theta = 1$.
• The trick is to factor out $\cos\theta$ from the numerator and $\sin\theta$ from the denominator, then bring $\cot\theta = \frac{\cos\theta}{\sin\theta}$ over a common denominator. The bracket simplifies to $2 - 2(\sin^2\theta+\cos^2\theta) = 2-2 = 0$.
• Examiners award marks for: correct factoring (1 mark), combining fractions correctly (1 mark), and using the Pythagorean identity to reach 0 (1 mark).
• Note: This is closely related to Exercise 8.3 Q4(vii), where $\frac{\sin\theta - 2\sin^3\theta}{2\cos^3\theta - \cos\theta} = \tan\theta$ — the same factoring technique applies.