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Prove that $\frac{1-\cos \theta}{1 + \cos \theta} = (\cot \theta - \operatorname{cosec} \theta)^2$
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LHS $= \frac{1-\cos \theta}{1 + \cos \theta} \times \frac{1-\cos \theta}{1-\cos \theta}$
$= \frac{(1-\cos \theta)^2}{1-\cos^2 \theta} = \frac{(1-\cos \theta)^2}{\sin^2 \theta}$
$= (\frac{1-\cos \theta}{\sin \theta})^2 = (\frac{1}{\sin \theta} - \frac{\cos \theta}{\sin \theta})^2$
$= (\operatorname{cosec} \theta - \cot \theta)^2 = \text{RHS}$
$= \frac{(1-\cos \theta)^2}{1-\cos^2 \theta} = \frac{(1-\cos \theta)^2}{\sin^2 \theta}$
$= (\frac{1-\cos \theta}{\sin \theta})^2 = (\frac{1}{\sin \theta} - \frac{\cos \theta}{\sin \theta})^2$
$= (\operatorname{cosec} \theta - \cot \theta)^2 = \text{RHS}$