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This section comprises Short Answer (SA) type questions of $3$ marks each.
Prove that : $\frac{\tan \theta - \cot \theta}{\sin \theta \cos \theta} = \sec^2 \theta - \operatorname{cosec}^2 \theta$
Prove that : $\frac{\tan \theta - \cot \theta}{\sin \theta \cos \theta} = \sec^2 \theta - \operatorname{cosec}^2 \theta$
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$LHS = \frac{\frac{\sin \theta}{\cos \theta} - \frac{\cos \theta}{\sin \theta}}{\sin \theta \cos \theta}$
$= \frac{\frac{\sin^2 \theta - \cos^2 \theta}{\sin \theta \cos \theta}}{\sin \theta \cos \theta}$
$= \frac{\sin^2 \theta - \cos^2 \theta}{\sin^2 \theta \cos^2 \theta}$
$= \frac{\sin^2 \theta}{\sin^2 \theta \cos^2 \theta} - \frac{\cos^2 \theta}{\sin^2 \theta \cos^2 \theta}$
$= \frac{1}{\cos^2 \theta} - \frac{1}{\sin^2 \theta}$
$= \sec^2 \theta - \operatorname{cosec}^2 \theta = RHS$
$= \frac{\frac{\sin^2 \theta - \cos^2 \theta}{\sin \theta \cos \theta}}{\sin \theta \cos \theta}$
$= \frac{\sin^2 \theta - \cos^2 \theta}{\sin^2 \theta \cos^2 \theta}$
$= \frac{\sin^2 \theta}{\sin^2 \theta \cos^2 \theta} - \frac{\cos^2 \theta}{\sin^2 \theta \cos^2 \theta}$
$= \frac{1}{\cos^2 \theta} - \frac{1}{\sin^2 \theta}$
$= \sec^2 \theta - \operatorname{cosec}^2 \theta = RHS$