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If $\sin \theta + \cos \theta = p$ and $\sec \theta + \cosec \theta = q$, then prove that $q (p^2 - 1) = 2p$.
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L.H.S = $q (p^2 - 1)$
$= (\sec \theta + \cosec \theta) [(\sin \theta + \cos \theta)^2 - 1]$
$= \left(\frac{1}{\cos \theta} + \frac{1}{\sin \theta}\right) [\sin^2\theta + \cos^2\theta + 2 \sin \theta \cos \theta - 1]$
$= \left(\frac{\sin\theta+\cos\theta}{\sin\theta\cos\theta}\right)[1+2 \sin \theta \cos \theta - 1]$
$= 2(\sin \theta+\cos \theta)$
$= 2p$
$= R.H.S$
$= (\sec \theta + \cosec \theta) [(\sin \theta + \cos \theta)^2 - 1]$
$= \left(\frac{1}{\cos \theta} + \frac{1}{\sin \theta}\right) [\sin^2\theta + \cos^2\theta + 2 \sin \theta \cos \theta - 1]$
$= \left(\frac{\sin\theta+\cos\theta}{\sin\theta\cos\theta}\right)[1+2 \sin \theta \cos \theta - 1]$
$= 2(\sin \theta+\cos \theta)$
$= 2p$
$= R.H.S$