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Prove that $\frac{\text{cosec}^2 \theta - \sec^2 \theta}{\text{cosec}^2 \theta + \sec^2 \theta} = \frac{3}{4}$, if $\tan \theta = \frac{1}{\sqrt{7}}$
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$\tan \theta = \frac{1}{\sqrt{7}}$
$\Rightarrow \sec^2\theta = \frac{8}{7}$ and $\text{cosec}^2\theta = 8$
$\therefore \text{LHS} = \frac{8 - \frac{8}{7}}{8 + \frac{8}{7}} = \frac{\frac{48}{7}}{\frac{64}{7}} = \frac{3}{4} = \text{RHS}$
$\Rightarrow \sec^2\theta = \frac{8}{7}$ and $\text{cosec}^2\theta = 8$
$\therefore \text{LHS} = \frac{8 - \frac{8}{7}}{8 + \frac{8}{7}} = \frac{\frac{48}{7}}{\frac{64}{7}} = \frac{3}{4} = \text{RHS}$