If $\sin \theta + \cos \theta = p$ and $\sec \theta + \operatorname{cosec} \theta = q,$ then prove that $q(p^2 - 1) = 2p$.

(N/A) Given that,$\sin \theta + \cos \theta = p$ ......$(i)$
And $\sec \theta + \operatorname{cosec} \theta = q$
$\Rightarrow \frac{1}{\cos \theta} + \frac{1}{\sin \theta} = q$ $\left[\because \sec \theta = \frac{1}{\cos \theta} \text{ and } \operatorname{cosec} \theta = \frac{1}{\sin \theta}\right]$
$\Rightarrow \frac{\sin \theta + \cos \theta}{\sin \theta \cdot \cos \theta} = q$
$\Rightarrow \frac{p}{\sin \theta \cdot \cos \theta} = q$ [from Eq. $(i)$]
$\Rightarrow \sin \theta \cdot \cos \theta = \frac{p}{q}$ ......$(ii)$
Now,squaring Eq. $(i)$ on both sides:
$(\sin \theta + \cos \theta)^2 = p^2$
$\Rightarrow (\sin^2 \theta + \cos^2 \theta) + 2 \sin \theta \cdot \cos \theta = p^2$ $\left[\because (a+b)^2 = a^2 + 2ab + b^2\right]$
$\Rightarrow 1 + 2 \sin \theta \cdot \cos \theta = p^2$ $\left[\because \sin^2 \theta + \cos^2 \theta = 1\right]$
Substitute $\sin \theta \cdot \cos \theta = \frac{p}{q}$ from Eq. $(ii)$:
$\Rightarrow 1 + 2 \left(\frac{p}{q}\right) = p^2$
$\Rightarrow 1 + \frac{2p}{q} = p^2$
Multiply both sides by $q$:
$\Rightarrow q + 2p = p^2 q$
$\Rightarrow 2p = p^2 q - q$
$\Rightarrow 2p = q(p^2 - 1)$
Hence,$q(p^2 - 1) = 2p$ is proved.