Explanation

(b)
Since,$\sec{\alpha,\ cosec\ \alpha\ }$are the roots of the equation
$x^{2} - px + q = 0$
$\therefore\sec{\alpha + cosec\ \alpha = p,\sec{\alpha \bullet cosec\ \alpha = q}}$
$\Rightarrow \frac{\sin{\alpha + \cos\alpha}}{\sin{\alpha\cos\alpha}} = p,\sin{\alpha\cos{\alpha = \frac{1}{q}}}$
$\Rightarrow \sin{\alpha + \cos{\alpha = \frac{p}{q}}}$
$\Rightarrow \sin^{2}{\alpha + \cos^{2}{\alpha + 2\sin{\alpha\cos{\alpha = \frac{p^{2}}{q^{2}}}}}}$
$\Rightarrow 1 + \frac{2}{q} = \frac{p^{2}}{q^{2}} \Rightarrow q(q + 2) = \ p^{2}$