Explanation

(b)
Since, $\tan{\alpha\ and\tan\beta}$ are the roots of the equation $x^{2} + ax + b = 0,$ then
$\tan{\alpha + \tan{\beta = - \frac{a}{1}}}$
and $\tan{\alpha\ \bullet \tan{\beta = b}}$
$\Rightarrow \frac{\sin\alpha}{\cos\alpha} + \frac{\sin\beta}{\cos\beta} = - \frac{a}{1}$
and $\ \frac{\sin\alpha\sin\beta\ }{\cos\alpha\ \cos\beta} = b$
$\therefore\sin^{2}{(\alpha + \beta) + a\sin(\alpha + \beta)}\cos{(\alpha + \beta) + b\cos^{2}(\alpha + \beta)}$
$= \ \cos^{2}{((\alpha + \beta)}\lbrack\tan^{2}{(\alpha + \beta) + b + a\tan{(\alpha + \beta)\rbrack}}$
$= \frac{\tan^{2}(\alpha + \beta) + b + a\tan(\alpha + \beta)\ }{1 + \tan^{2}(\alpha + \beta)}$
$= \frac{\frac{a}{b - 1}\left( \ a + \frac{a}{b - 1} \right)}{1 + \frac{a^{2}}{(b - 1)^{2}}} = b$