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Prove that : $\sqrt{\frac{1-\sin \theta}{1 + \sin \theta}} = \sec \theta - \tan \theta$
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LHS $= \sqrt{\frac{(1 - \sin \theta)}{(1 + \sin \theta)} \times \frac{(1 - \sin \theta)}{(1- \sin \theta)}}$ (1/2 Mark)
$= \sqrt{\frac{(1 - \sin \theta)^2}{(1- \sin^2 \theta)}}$ (1/2 Mark)
$= \sqrt{\frac{(1 - \sin \theta)^2}{\cos^2 \theta}}$ (1/2 Mark)
$= \frac{(1 - \sin \theta)}{\cos \theta} = \sec\theta - \tan\theta = RHS$ (1/2 Mark)
$= \sqrt{\frac{(1 - \sin \theta)^2}{(1- \sin^2 \theta)}}$ (1/2 Mark)
$= \sqrt{\frac{(1 - \sin \theta)^2}{\cos^2 \theta}}$ (1/2 Mark)
$= \frac{(1 - \sin \theta)}{\cos \theta} = \sec\theta - \tan\theta = RHS$ (1/2 Mark)