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If $\frac{\sec \alpha}{\operatorname{cosec} \beta} = p$ and $\frac{\tan \alpha}{\operatorname{cosec} \beta} = q$, then prove that $(p^2 – q^2) \sec^2 \alpha = p^2$.
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Sol. LHS $= (p^2 - q^2) \sec^2 \alpha$ (1/2 Mark)
$= (\frac{\sec^2 \alpha}{\operatorname{cosec}^2 \beta} - \frac{\tan^2 \alpha}{\operatorname{cosec}^2 \beta}) \times \sec^2 \alpha$ (1 Mark)
$= (\frac{\sec^2 \alpha - \tan^2 \alpha}{\operatorname{cosec}^2 \beta}) \times \sec^2 \alpha$ (1 Mark)
$= (\frac{1}{\operatorname{cosec}^2 \beta}) \times \sec^2 \alpha$ (1/2 Mark)
$= p^2 = \text{RHS}$
$= (\frac{\sec^2 \alpha}{\operatorname{cosec}^2 \beta} - \frac{\tan^2 \alpha}{\operatorname{cosec}^2 \beta}) \times \sec^2 \alpha$ (1 Mark)
$= (\frac{\sec^2 \alpha - \tan^2 \alpha}{\operatorname{cosec}^2 \beta}) \times \sec^2 \alpha$ (1 Mark)
$= (\frac{1}{\operatorname{cosec}^2 \beta}) \times \sec^2 \alpha$ (1/2 Mark)
$= p^2 = \text{RHS}$