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Prove that : $\frac{1+\sin \theta}{1-\sin \theta} - \frac{1-\sin \theta}{1+\sin \theta} = 4 \tan \theta \sec \theta$
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Sol. LHS $= \frac{(1+\sin\theta)^2-(1-\sin \theta)^2}{(1+\sin \theta) (1-\sin \theta)}$ (2 Marks)
$= \frac{4 \sin \theta}{1-\sin^2\theta}$ (1 Mark)
$= \frac{4 \sin \theta}{\cos^2\theta}$ (1 Mark)
$= 4 \tan \theta \sec \theta = \text{RHS}$ (1 Mark)
$= \frac{4 \sin \theta}{1-\sin^2\theta}$ (1 Mark)
$= \frac{4 \sin \theta}{\cos^2\theta}$ (1 Mark)
$= 4 \tan \theta \sec \theta = \text{RHS}$ (1 Mark)