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