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