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