Mathematics — CBSE Class 10 board question

LHS $= \dfrac{\cosec\theta}{\cosec\theta - 1} + \dfrac{\cosec\theta}{\cosec\theta + 1}$

Taking LCM:

$$= \frac{\cosec\theta(\cosec\theta + 1) + \cosec\theta(\cosec\theta - 1)}{(\cosec\theta - 1)(\cosec\theta + 1)}$$

$$= \frac{\cosec^2\theta + \cosec\theta + \cosec^2\theta - \cosec\theta}{\cosec^2\theta - 1}$$

$$= \frac{2\cosec^2\theta}{\cosec^2\theta - 1}$$

Using identity $\cosec^2\theta - 1 = \cot^2\theta$:

$$= \frac{2\cosec^2\theta}{\cot^2\theta} = 2 \times \frac{1}{\sin^2\theta} \times \frac{\sin^2\theta}{\cos^2\theta} = \frac{2}{\cos^2\theta} = 2\sec^2\theta = \textbf{RHS}$$

Hence proved.

Source: Chapter 8, Section 8.4 Trigonometric Identities

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• Key identities used: $\cosec^2\theta - 1 = \cot^2\theta$ and $\sec\theta = \dfrac{1}{\cos\theta}$, $\cot\theta = \dfrac{\cos\theta}{\sin\theta}$.
• The standard move is to take LCM first, then simplify the numerator, and apply the Pythagorean identity in the denominator.
• Examiners award marks stepwise: LCM step (1 mark), simplification to $\dfrac{2\cosec^2\theta}{\cot^2\theta}$ (1 mark), final conversion to $2\sec^2\theta$ (1 mark). Show each step clearly.