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Prove that : $\frac{\tan \theta}{1-\cot \theta} + \frac{\cot \theta}{1-\tan \theta} = 1 + \sec \theta \cosec \theta$
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$LHS = \frac{\sin \theta / \cos \theta}{1-\cos \theta / \sin \theta} + \frac{\cos \theta / \sin \theta}{1-\sin \theta / \cos \theta}$
$= \frac{\sin^2 \theta}{\cos \theta (\sin \theta - \cos \theta)} + \frac{\cos^2 \theta}{\sin \theta (\cos \theta - \sin \theta)}$
$= \frac{\sin^3 \theta - \cos^3 \theta}{\sin \theta \cos \theta (\sin \theta - \cos \theta)}$
$= \frac{(\sin \theta - \cos \theta)(\sin^2 \theta + \cos^2 \theta + \sin \theta \cos \theta)}{\sin \theta \cos \theta (\sin \theta - \cos \theta)}$
$= \frac{1 + \sin \theta \cos \theta}{\sin \theta \cos \theta}$
$= \frac{1}{\sin \theta \cos \theta} + 1$
$= 1 + \cosec \theta \sec \theta = RHS$
$= \frac{\sin^2 \theta}{\cos \theta (\sin \theta - \cos \theta)} + \frac{\cos^2 \theta}{\sin \theta (\cos \theta - \sin \theta)}$
$= \frac{\sin^3 \theta - \cos^3 \theta}{\sin \theta \cos \theta (\sin \theta - \cos \theta)}$
$= \frac{(\sin \theta - \cos \theta)(\sin^2 \theta + \cos^2 \theta + \sin \theta \cos \theta)}{\sin \theta \cos \theta (\sin \theta - \cos \theta)}$
$= \frac{1 + \sin \theta \cos \theta}{\sin \theta \cos \theta}$
$= \frac{1}{\sin \theta \cos \theta} + 1$
$= 1 + \cosec \theta \sec \theta = RHS$