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Prove that :
$\frac{\sec^3 \theta}{\sec^2 \theta - 1} + \frac{\operatorname{cosec}^3 \theta}{\operatorname{cosec}^2 \theta - 1} = \sec \theta \cdot \operatorname{cosec} \theta (\sec \theta + \operatorname{cosec} \theta)$
$\frac{\sec^3 \theta}{\sec^2 \theta - 1} + \frac{\operatorname{cosec}^3 \theta}{\operatorname{cosec}^2 \theta - 1} = \sec \theta \cdot \operatorname{cosec} \theta (\sec \theta + \operatorname{cosec} \theta)$
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Sol. LHS $= \frac{\sec^3 \theta}{(\sec^2 \theta - 1)} + \frac{\operatorname{cosec}^3 \theta}{(\operatorname{cosec}^2 \theta - 1)}$ (1 Mark)
$= \frac{\sec^3 \theta}{\tan^2 \theta} + \frac{\operatorname{cosec}^3 \theta}{\cot^2 \theta}$ (1/2 Mark)
$= \frac{1}{\cos^3 \theta} \times \frac{\cos^2 \theta}{\sin^2 \theta} + \frac{1}{\sin^3 \theta} \times \frac{\sin^2 \theta}{\cos^2 \theta}$ (1/2 Mark)
$= \frac{1}{\cos \theta \sin^2 \theta} + \frac{1}{\sin \theta \cos^2 \theta}$ (1/2 Mark)
$= \frac{1}{\sin \theta \cos \theta} [\frac{1}{\sin \theta} + \frac{1}{\cos \theta}]$ (1/2 Mark)
$= \sec \theta \cdot \operatorname{cosec} \theta (\sec \theta + \operatorname{cosec} \theta) = \text{RHS}$ (1/2 Mark)
$= \frac{\sec^3 \theta}{\tan^2 \theta} + \frac{\operatorname{cosec}^3 \theta}{\cot^2 \theta}$ (1/2 Mark)
$= \frac{1}{\cos^3 \theta} \times \frac{\cos^2 \theta}{\sin^2 \theta} + \frac{1}{\sin^3 \theta} \times \frac{\sin^2 \theta}{\cos^2 \theta}$ (1/2 Mark)
$= \frac{1}{\cos \theta \sin^2 \theta} + \frac{1}{\sin \theta \cos^2 \theta}$ (1/2 Mark)
$= \frac{1}{\sin \theta \cos \theta} [\frac{1}{\sin \theta} + \frac{1}{\cos \theta}]$ (1/2 Mark)
$= \sec \theta \cdot \operatorname{cosec} \theta (\sec \theta + \operatorname{cosec} \theta) = \text{RHS}$ (1/2 Mark)