Mathematics — CBSE Class 10 board question

LHS $= \dfrac{\sec^3\theta}{\sec^2\theta - 1} + \dfrac{\csc^3\theta}{\csc^2\theta - 1}$

Using identities: $\sec^2\theta - 1 = \tan^2\theta$ and $\csc^2\theta - 1 = \cot^2\theta$

$$= \frac{\sec^3\theta}{\tan^2\theta} + \frac{\csc^3\theta}{\cot^2\theta}$$

$$= \sec^3\theta \cdot \frac{\cos^2\theta}{\sin^2\theta} + \csc^3\theta \cdot \frac{\sin^2\theta}{\cos^2\theta}$$

$$= \frac{1}{\cos^3\theta} \cdot \frac{\cos^2\theta}{\sin^2\theta} + \frac{1}{\sin^3\theta} \cdot \frac{\sin^2\theta}{\cos^2\theta}$$

$$= \frac{1}{\cos\theta \cdot \sin^2\theta} + \frac{1}{\sin\theta \cdot \cos^2\theta}$$

$$= \frac{1}{\sin\theta\cos\theta}\left(\frac{1}{\sin\theta} + \frac{1}{\cos\theta}\right)$$

$$= \sec\theta\cdot\csc\theta\,(\csc\theta + \sec\theta) = \textbf{RHS} \quad \blacksquare$$

Source: Chapter 8, Section 8.4 — Trigonometric Identities

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• The key step is substituting $\sec^2\theta - 1 = \tan^2\theta$ and $\csc^2\theta - 1 = \cot^2\theta$ (standard identities from the textbook).
• Then convert $\tan^2\theta$ and $\cot^2\theta$ into $\sin/\cos$ form to simplify each fraction.
• Taking $\frac{1}{\sin\theta\cos\theta}$ common at the end gives the RHS directly.
• Examiners award marks at each conversion step, so show every line clearly.