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