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