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