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If $\tan \theta = \frac{1}{\sqrt{7}}$, then show that $\frac{\text{cosec}^2 \theta - \sec^2 \theta}{\text{cosec}^2\theta+ \sec^2 \theta} = \frac{3}{4}$
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$$\begin{aligned}& \sec^2 \theta = 1 + \frac{1}{7} = \frac{8}{7} \\ & \cot \theta = \sqrt{7} \Rightarrow \text{cosec}^2 \theta = 1 + 7 = 8 \\ & \therefore \text{LHS} = \frac{8 - \frac{8}{7}}{8 + \frac{8}{7}} = \frac{\frac{48}{7}}{\frac{64}{7}} \\ & = \frac{3}{4} = \text{RHS}\end{aligned}$$