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Prove that : $\frac{\tan^3 \theta}{1+\tan^2 \theta} + \frac{\cot^3 \theta}{1 + \cot^2 \theta} = \sec \theta \cosec \theta - 2 \sin \theta \cos \theta$
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LHS $= \frac{\tan^3 \theta}{\sec^2 \theta} + \frac{\cot^3 \theta}{\cosec^2 \theta}$
$= \frac{\sin^3 \theta}{\cos \theta} + \frac{\cos^3 \theta}{\sin \theta}$
$= \frac{\sin^4 \theta + \cos^4 \theta}{\sin \theta \cos \theta}$
$= \frac{(\sin^2 \theta + \cos^2 \theta)^2 - 2\sin^2 \theta \cos^2 \theta}{\sin \theta \cos \theta}$
$= \frac{1 - 2\sin^2 \theta \cos^2 \theta}{\sin \theta \cos \theta} = \frac{1}{\sin \theta \cos \theta} - \frac{2\sin^2 \theta \cos^2 \theta}{\sin \theta \cos \theta}$
$= \cosec \theta \sec \theta - 2 \sin \theta \cos \theta = \text{RHS}$
$= \frac{\sin^3 \theta}{\cos \theta} + \frac{\cos^3 \theta}{\sin \theta}$
$= \frac{\sin^4 \theta + \cos^4 \theta}{\sin \theta \cos \theta}$
$= \frac{(\sin^2 \theta + \cos^2 \theta)^2 - 2\sin^2 \theta \cos^2 \theta}{\sin \theta \cos \theta}$
$= \frac{1 - 2\sin^2 \theta \cos^2 \theta}{\sin \theta \cos \theta} = \frac{1}{\sin \theta \cos \theta} - \frac{2\sin^2 \theta \cos^2 \theta}{\sin \theta \cos \theta}$
$= \cosec \theta \sec \theta - 2 \sin \theta \cos \theta = \text{RHS}$