Given that, $\sin \theta+\cos \theta=p$ ……$(i)$

and $\sec \theta+\operatorname{cosec} \theta=q$

$\Rightarrow \quad \frac{1}{\cos \theta}+\frac{1}{\sin \theta}=q$ $\left[\because \sec \theta=\frac{1}{\cos \theta}\right.$ and $\left.\operatorname{cosec} \theta=\frac{1}{\sin \theta}\right]$

$\Rightarrow \quad \frac{\sin \theta+\cos \theta}{\sin \theta \cdot \cos \theta}=q$

$\Rightarrow \quad \frac{p}{\sin \theta \cdot \cos \theta}=q$ [from Eq. $(i)$]

$\Rightarrow \quad \sin \theta \cdot \cos \theta=\frac{p}{q}$ [from Eq. $(i)$] …..$(ii)$

$\sin \theta+\cos \theta=p$

On squaring both sides. we get

$(\sin \theta+\cos \theta)^{2}=p^{2}$

$\Rightarrow \quad\left(\sin ^{2} \theta+\cos ^{2} \theta\right)+2 \sin \theta \cdot \cos \theta=p^{2} \quad\left[\because(a+b)^{2}=a^{2}+2 a b+b^{2}\right]$

$\Rightarrow \quad 1+2 \sin \theta \cdot \cos \theta=p^{2} \quad\left[\because \sin ^{2} \theta+\cos ^{2} \theta=1\right]$

$\Rightarrow$ $1+2 \cdot \frac{p}{q}=p^{2}$ [from Eq.$(ii)$]

$\Rightarrow \quad q+2 p=p^{2} q \Rightarrow 2 p=p^{2} q-q$

$\Rightarrow$ $q\left(p^{2}-1\right)=2 p$ Hence proved.