If $\tan \mathrm{\alpha },\tan \mathrm{\beta }$ are the roots of the equation x² + px + q = 0, then the value of ${\sin }^2(\mathrm{\alpha }+\mathrm{\beta })+\mathrm{psin}(\mathrm{\alpha }+\mathrm{\beta })\cos (\mathrm{\alpha }+\mathrm{\beta })+{\mathrm{qcos}}^2(\mathrm{\alpha }+\mathrm{\beta })$ is

$\begin{array}{c}\tan \mathrm{\alpha }+\tan \mathrm{\beta }=-\mathrm{p}\\ \tan \mathrm{\alpha tan}\mathrm{\beta }=\mathrm{q}\\ \tan (\mathrm{\alpha }+\mathrm{\beta })=\frac{-\mathrm{p}}{1-\mathrm{q}}=\frac{\mathrm{p}}{\mathrm{q}-1}\\ \frac{1}{1+{\tan }^2(\mathrm{\alpha }+\mathrm{\beta })}\left[{\tan }^2(\mathrm{\alpha }+\mathrm{\beta })+\mathrm{ptan}(\mathrm{\alpha }+\mathrm{\beta })+\mathrm{q}\right]\\ =\frac{1}{1+\frac{{\mathrm{p}}^2}{(\mathrm{q}-1{)}^2}}\left[\frac{{\mathrm{p}}^2}{(\mathrm{q}-1{)}^2}+\frac{{\mathrm{p}}^2}{(\mathrm{q}-1)}+\mathrm{q}\right]\\ =\frac{1}{(\mathrm{q}-1{)}^2+{\mathrm{p}}^2}\left[{\mathrm{p}}^2+{\mathrm{p}}^2(\mathrm{q}-1)+\mathrm{q}(\mathrm{q}-1{)}^2\right]\\ =\frac{1}{{\mathrm{p}}^2+(\mathrm{q}-1{)}^2}{\left[{\mathrm{p}}^2\mathrm{q}+\mathrm{q}(\mathrm{q}-1{)}^2\right]}^4\\ =\mathrm{q}\left[\frac{{\mathrm{p}}^2+(\mathrm{q}-1{)}^2}{{\mathrm{p}}^2+(\mathrm{q}-1{)}^2}\right]=\mathrm{q}\end{array}$