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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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Sol. LHS = $\frac{\frac{\sin\theta}{\cos\theta}}{1-\frac{\cos\theta}{\sin\theta}} + \frac{\frac{\cos\theta}{\sin\theta}}{1-\frac{\sin\theta}{\cos\theta}}$
$= \frac{\sin^2\theta}{\cos\theta(\sin\theta-\cos\theta)} - \frac{\cos^2\theta}{\sin\theta(\sin\theta-\cos\theta)}$
$= \frac{1}{(\sin\theta-\cos\theta)} \left[\frac{\sin^3\theta - \cos^3\theta}{\sin\theta\cos\theta}\right]$
$= \frac{1}{(\sin\theta-\cos\theta)} \times \frac{(\sin\theta-\cos\theta)(\sin^2\theta + \cos^2\theta + \sin\theta\cos\theta)}{\sin\theta\cos\theta}$
$= \frac{1 + \sin\theta\cos\theta}{\sin\theta\cos\theta}$
$= \frac{1}{\sin\theta\cos\theta} + 1$
$= 1 + \sec\theta \cosec\theta = \text{RHS}$
$= \frac{\sin^2\theta}{\cos\theta(\sin\theta-\cos\theta)} - \frac{\cos^2\theta}{\sin\theta(\sin\theta-\cos\theta)}$
$= \frac{1}{(\sin\theta-\cos\theta)} \left[\frac{\sin^3\theta - \cos^3\theta}{\sin\theta\cos\theta}\right]$
$= \frac{1}{(\sin\theta-\cos\theta)} \times \frac{(\sin\theta-\cos\theta)(\sin^2\theta + \cos^2\theta + \sin\theta\cos\theta)}{\sin\theta\cos\theta}$
$= \frac{1 + \sin\theta\cos\theta}{\sin\theta\cos\theta}$
$= \frac{1}{\sin\theta\cos\theta} + 1$
$= 1 + \sec\theta \cosec\theta = \text{RHS}$