Mathematics — CBSE Class 10 board question

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

Taking LCM:

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

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

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

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

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

Hence proved. $\blacksquare$

Source: Chapter 8, Section 8.4 Trigonometric Identities

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• Key identities used: $\cosec^2 A - 1 = \cot^2 A$ and $\dfrac{\cosec^2 A}{\cot^2 A} = \dfrac{1/\sin^2 A}{\cos^2 A/\sin^2 A} = \sec^2 A$.
• Always take LCM first, simplify the numerator, then apply the Pythagorean identity to the denominator. Examiners award 1 mark for the LCM step, 1 mark for applying the identity, and 1 mark for reaching RHS.