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Prove that : $\sqrt{\sec^2 \theta + \text{cosec}^2 \theta} = \tan \theta + \cot \theta$
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LHS $= \sqrt{\frac{1}{\cos^2 \theta} + \frac{1}{\sin^2 \theta}}$ ($1/2$)
$= \sqrt{\frac{\sin^2 \theta + \cos^2 \theta}{\sin^2 \theta \cos^2 \theta}}$ ($1/2$)
$= \frac{1}{\sin \theta \cos \theta}$ ($1$)
$= \frac{\sin^2 \theta + \cos^2 \theta}{\sin \theta \cos \theta}$ ($1/2$)
$= \frac{\sin^2 \theta}{\sin \theta \cos \theta} + \frac{\cos^2 \theta}{\sin \theta \cos \theta}$
$= \tan \theta + \cot \theta = \text{RHS}$ ($1/2$)
$= \sqrt{\frac{\sin^2 \theta + \cos^2 \theta}{\sin^2 \theta \cos^2 \theta}}$ ($1/2$)
$= \frac{1}{\sin \theta \cos \theta}$ ($1$)
$= \frac{\sin^2 \theta + \cos^2 \theta}{\sin \theta \cos \theta}$ ($1/2$)
$= \frac{\sin^2 \theta}{\sin \theta \cos \theta} + \frac{\cos^2 \theta}{\sin \theta \cos \theta}$
$= \tan \theta + \cot \theta = \text{RHS}$ ($1/2$)