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