If tan A and tan B are the roots of the quadratic equation x² – px + q = 0, then value of sin² (A + B) is

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

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

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

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

answer is D.

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Detailed Solution

We have tan A + tan B = p, and tan A tan B = q.

$∴ \tan (A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} = \frac{p}{1 - q}$

Now, $\sin^{2} (A + B) = \frac{\tan^{2} (A + B)}{\sec^{2} (A + B)}$

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

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