Given that $t a n A, t a n B$ are the roots of the equation $x^{2} - p x + q = 0$ , then the value of $s i n^{2} (A + B)$ is

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

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\frac{p^{2}}{p^{2} + q^{2}}

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

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\frac{p^{2}}{(p + q)^{2}}

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Solution

The correct option is A $\frac{p^{2}}{p^{2} + (1 - q)^{2}}$
From the given condition we have
$t a n A + t a n B = p, t a n A t a n B = q$
So that $t a n (A + B) = \frac{t a n A + t a n B}{1 - t a n A t a n B} = \frac{p}{1 - q}$
And $s i n^{2} (A + B) = \frac{t a n^{2} (A + B)}{1 + t a n^{2} (A + B)} = \frac{[\frac{p^{2}}{(1 - q)^{2}}]}{[1 + \frac{p^{2}}{(1 - q)^{2}}]} = \frac{p^{2}}{p^{2} + (1 - q)^{2}}$

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