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If $\cot \theta + \cos \theta = p$ and $\cot \theta-\cos \theta = q$, prove that $p^2 – q^2 = 4\sqrt{pq}$
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LHS = $p^2 – q^2$
$= (\cot \theta + \cos \theta)^2 – (\cot \theta – \cos \theta)^2$
$= [(\cot \theta + \cos \theta) + (\cot \theta – \cos \theta)][(\cot \theta + \cos \theta) – (\cot \theta - \cos \theta)]$
$= 2 \cot \theta \times 2 \cos \theta = 4 \cot \theta \cos \theta$
RHS = $4\sqrt{pq}$
$= 4\sqrt{(\cot \theta + \cos \theta) (\cot \theta – \cos \theta)}$
$= 4\sqrt{\cot^2\theta - \cos^2\theta}$
$= 4\sqrt{\cos^2\theta(\text{cosec}^2\theta – 1)}$
$= 4\sqrt{\cos^2\theta \times \cot^2\theta}$
$= 4 \cot \theta \cos \theta$
$\therefore$ LHS = RHS
$= (\cot \theta + \cos \theta)^2 – (\cot \theta – \cos \theta)^2$
$= [(\cot \theta + \cos \theta) + (\cot \theta – \cos \theta)][(\cot \theta + \cos \theta) – (\cot \theta - \cos \theta)]$
$= 2 \cot \theta \times 2 \cos \theta = 4 \cot \theta \cos \theta$
RHS = $4\sqrt{pq}$
$= 4\sqrt{(\cot \theta + \cos \theta) (\cot \theta – \cos \theta)}$
$= 4\sqrt{\cot^2\theta - \cos^2\theta}$
$= 4\sqrt{\cos^2\theta(\text{cosec}^2\theta – 1)}$
$= 4\sqrt{\cos^2\theta \times \cot^2\theta}$
$= 4 \cot \theta \cos \theta$
$\therefore$ LHS = RHS