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Prove that: $\left( \frac{1}{\cos \theta} - \cos \theta \right) \left( \frac{1}{\sin \theta} - \sin \theta \right) = \frac{1}{\tan \theta + \cot \theta}$
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LHS = $\left( \frac{1}{\cos \theta} - \cos \theta \right) \left( \frac{1}{\sin \theta} - \sin \theta \right)$
$= \left( \frac{1 - \cos^2 \theta}{\cos \theta} \right) \left( \frac{1 - \sin^2 \theta}{\sin \theta} \right)$
$= \frac{\sin^2 \theta}{\cos \theta} \times \frac{\cos^2 \theta}{\sin \theta}$
$= \sin \theta \cos \theta$
RHS = $\frac{1}{\tan \theta + \cot \theta} = \frac{1}{\frac{\sin \theta}{\cos \theta} + \frac{\cos \theta}{\sin \theta}}$
$= \frac{1}{\frac{\sin^2 \theta + \cos^2 \theta}{\cos \theta \sin \theta}}$
$= \frac{1}{\frac{1}{\sin \theta \cos \theta}}$
$= \sin \theta \cos \theta$
$\therefore$ LHS = RHS
$= \left( \frac{1 - \cos^2 \theta}{\cos \theta} \right) \left( \frac{1 - \sin^2 \theta}{\sin \theta} \right)$
$= \frac{\sin^2 \theta}{\cos \theta} \times \frac{\cos^2 \theta}{\sin \theta}$
$= \sin \theta \cos \theta$
RHS = $\frac{1}{\tan \theta + \cot \theta} = \frac{1}{\frac{\sin \theta}{\cos \theta} + \frac{\cos \theta}{\sin \theta}}$
$= \frac{1}{\frac{\sin^2 \theta + \cos^2 \theta}{\cos \theta \sin \theta}}$
$= \frac{1}{\frac{1}{\sin \theta \cos \theta}}$
$= \sin \theta \cos \theta$
$\therefore$ LHS = RHS