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(A) From the given quadratic equation $x^2 - px + q = 0$,we have the sum and product of roots as:$	an \alpha + 	an \beta = p$ $(i)$$	an \alpha 	an

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$\frac{p^2}{p^2 + (1 - q)^2}$

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(A) From the given quadratic equation $x^2 - px + q = 0$,we have the sum and product of roots as:$	an \alpha + 	an \beta = p$ $(i)$$	an \alpha 	an \beta = q$ $(ii)$Using the formula for $	an(\alpha + \beta)$:$	an(\alpha + \beta) = \frac{	an \alpha + 	an \beta}{1 - 	an \alpha 	an \beta} = \frac{p}{1 - q}$We know that $\sin^2 	heta = \frac{	an^2 	heta}{1 + 	an^2 	heta}$.Substituting $	heta = \alpha + \beta$:$\sin^2(\alpha + \beta) = \frac{	an^2(\alpha + \beta)}{1 + 	an^2(\alpha + \beta)}$$= \frac{(\frac{p}{1 - q})^2}{1 + (\frac{p}{1 - q})^2}$$= \frac{\frac{p^2}{(1 - q)^2}}{\frac{(1 - q)^2 + p^2}{(1 - q)^2}}$$= \frac{p^2}{p^2 + (1 - q)^2}$.

Given that $\tan \alpha$ and $\tan \beta$ are the roots of $x^2 - px + q = 0$,then the value of $\sin^2(\alpha + \beta)$ is:

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