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If $\sec \theta + \tan \theta = m$, show that $\frac{m^2 - 1}{m^2 + 1} = \sin \theta$.
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L.H.S. $= \frac{m^2 - 1}{m^2 + 1}$ (I) (1 Mark)
$= \frac{(\sec \theta+\tan \theta)^2 - (\sec^2 \theta - \tan^2 \theta)}{(\sec \theta+\tan \theta)^2 + (\sec^2 \theta - \tan^2 \theta)}$ (II) (1 Mark)
$= \frac{(\sec \theta+\tan \theta) (\sec \theta+\tan \theta - \sec \theta+\tan \theta)}{(\sec \theta+\tan \theta) (\sec \theta+\tan \theta + \sec \theta - \tan \theta)}$ (III) (1/2 Mark)
$= \frac{\tan \theta}{\sec \theta}$ (IV) (1/2 Mark)
$= \sin \theta = \text{R.H.S.}$
$= \frac{(\sec \theta+\tan \theta)^2 - (\sec^2 \theta - \tan^2 \theta)}{(\sec \theta+\tan \theta)^2 + (\sec^2 \theta - \tan^2 \theta)}$ (II) (1 Mark)
$= \frac{(\sec \theta+\tan \theta) (\sec \theta+\tan \theta - \sec \theta+\tan \theta)}{(\sec \theta+\tan \theta) (\sec \theta+\tan \theta + \sec \theta - \tan \theta)}$ (III) (1/2 Mark)
$= \frac{\tan \theta}{\sec \theta}$ (IV) (1/2 Mark)
$= \sin \theta = \text{R.H.S.}$