If α and β are the roots of the equation x 2 + p x + 2 = 0 and 1 α and 1 β are the roots…

$\alpha , \beta$ are roots of $x^2+px+2=0$

$\Rightarrow \alpha +\beta =-p, \alpha \beta =2$

$\frac{1}{\alpha },\frac{1}{\beta }$ are roots of  $2x^2+2qx+1=0$ 

$\therefore \alpha , \beta$ are roots of $x^2+2qx+2=0$

 $\Rightarrow \alpha +\beta =-2q, \alpha \beta =2$

$\therefore p=2q ...\left(\mathrm{i}\right)$

$\Rightarrow \left(\alpha -\frac{1}{\alpha }\right)\left(\beta -\frac{1}{\beta }\right)\left(\alpha +\frac{1}{\beta }\right)\left(\beta +\frac{1}{\alpha }\right)=\left\{\alpha \beta -\frac{\alpha }{\beta }-\frac{\beta }{\alpha }+\frac{1}{\alpha \beta }\right\}·\left\{\alpha \beta +\frac{1}{\alpha \beta }+2\right\}$

$\Rightarrow \left\{2-\left(\frac{{\left(\alpha +\beta \right)}^2-2\alpha \beta }{\alpha \beta }\right)+\frac{1}{2}\right\}·\left\{2+\frac{1}{2}+2\right\}$

$\Rightarrow \left\{\frac{5}{2}-\left(\frac{p^2-4}{2}\right)\right\}·\left\{\frac{9}{2}\right\}=\frac{9}{4}\left(9-p^2\right)$ or, $\frac{9}{4}\left(9-4q^2\right)$