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(C) Given that $	an \alpha$ and $	an \beta$ are the roots of $x^2+px+q=0$. From the relation between roots and coefficients,we have $	an \alpha + \

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$q$

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(C) Given that $	an \alpha$ and $	an \beta$ are the roots of $x^2+px+q=0$. From the relation between roots and coefficients,we have $	an \alpha + 	an \beta = -p$ and $	an \alpha 	an \beta = q$. Let $E = \sin^2(\alpha+\beta) + p\cos(\alpha+\beta)\sin(\alpha+\beta) + q\cos^2(\alpha+\beta)$. Dividing the expression by $\cos^2(\alpha+\beta)$,we get $E = \cos^2(\alpha+\beta) [	an^2(\alpha+\beta) + p	an(\alpha+\beta) + q]$. We know $	an(\alpha+\beta) = \frac{	an \alpha + 	an \beta}{1 - 	an \alpha 	an \beta} = \frac{-p}{1-q} = \frac{p}{q-1}$. Substituting this into the expression inside the bracket: $	an^2(\alpha+\beta) + p	an(\alpha+\beta) + q = \frac{p^2}{(q-1)^2} + p(\frac{p}{q-1}) + q = \frac{p^2 + p^2(q-1) + q(q-1)^2}{(q-1)^2} = \frac{p^2 + p^2q - p^2 + q(q^2-2q+1)}{(q-1)^2} = \frac{p^2q + q^3 - 2q^2 + q}{(q-1)^2} = \frac{q(p^2 + q^2 - 2q + 1)}{(q-1)^2} = \frac{q((q-1)^2 + p^2)}{(q-1)^2}$. Now,$\cos^2(\alpha+\beta) = \frac{1}{1+	an^2(\alpha+\beta)} = \frac{1}{1+\frac{p^2}{(q-1)^2}} = \frac{(q-1)^2}{(q-1)^2+p^2}$. Thus,$E = \frac{(q-1)^2}{(q-1)^2+p^2} 	imes \frac{q((q-1)^2 + p^2)}{(q-1)^2} = q$.

If $\tan \alpha$ and $\tan \beta$ are the roots of the equation $x^2+px+q=0$,then the value of $\sin^2(\alpha+\beta)+p\cos(\alpha+\beta)\sin(\alpha+\beta)+q\cos^2(\alpha+\beta)$ is

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