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Prove that : $\frac{\sin \theta}{1 + \cos \theta} + \frac{1 + \cos \theta}{\sin \theta} = 2 \operatorname{cosec} \theta$
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LHS = $\frac{\sin^2 \theta + (1 + \cos \theta)^2}{\sin \theta (1 + \cos \theta)}$ (1 Mark)
$= \frac{\sin^2 \theta + 1 + \cos^2 \theta + 2\cos \theta}{\sin \theta (1 + \cos \theta)}$ (1 Mark)
$= \frac{2 + 2\cos \theta}{\sin \theta (1 + \cos \theta)}$ (1/2 Mark)
$= \frac{2(1 + \cos \theta)}{\sin \theta (1 + \cos \theta)} = \frac{2}{\sin \theta} = 2 \operatorname{cosec} \theta = \text{RHS}$ (1/2 Mark)
$= \frac{\sin^2 \theta + 1 + \cos^2 \theta + 2\cos \theta}{\sin \theta (1 + \cos \theta)}$ (1 Mark)
$= \frac{2 + 2\cos \theta}{\sin \theta (1 + \cos \theta)}$ (1/2 Mark)
$= \frac{2(1 + \cos \theta)}{\sin \theta (1 + \cos \theta)} = \frac{2}{\sin \theta} = 2 \operatorname{cosec} \theta = \text{RHS}$ (1/2 Mark)