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If $4 \tan A = 3$, then find the value of $\frac{\text{cosec}^2 A + 1}{\text{cosec}^2 A - 1}$.
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$4 \tan A = 3$
$\tan A = \frac{3}{4}$ ($\frac{1}{2}$ Mark)
$\Rightarrow \cot A = \frac{4}{3}$ ($\frac{1}{2}$ Mark)
$\therefore \text{cosec}^2 A = 1 + (\frac{4}{3})^2 = 1 + \frac{16}{9} = \frac{25}{9}$ (1 Mark)
$\frac{\text{cosec}^2 A + 1}{\text{cosec}^2 A - 1} = \frac{\frac{25}{9} + 1}{\frac{25}{9} - 1} = \frac{\frac{34}{9}}{\frac{16}{9}} = \frac{34}{16}$ or $\frac{17}{8}$ (1 Mark)
$\tan A = \frac{3}{4}$ ($\frac{1}{2}$ Mark)
$\Rightarrow \cot A = \frac{4}{3}$ ($\frac{1}{2}$ Mark)
$\therefore \text{cosec}^2 A = 1 + (\frac{4}{3})^2 = 1 + \frac{16}{9} = \frac{25}{9}$ (1 Mark)
$\frac{\text{cosec}^2 A + 1}{\text{cosec}^2 A - 1} = \frac{\frac{25}{9} + 1}{\frac{25}{9} - 1} = \frac{\frac{34}{9}}{\frac{16}{9}} = \frac{34}{16}$ or $\frac{17}{8}$ (1 Mark)