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