Description for Correct answer:
We have, \( \Large x^{2}-px+q=0 \)

Here, \( \Large \tan \alpha + \tan \beta = P \) ...(i)

\( \Large \tan \alpha + \tan \beta = q \) ...(ii)

Hence, \( \Large \tan \left( \alpha + \beta \right) = \frac{\tan \alpha + \tan \beta }{1-\tan \alpha \tan \beta } = \frac{P}{1-q} \) ...(iii)

\( \Large \therefore \sin^{2} \left( \alpha + \beta \right) = \frac{1-\cos\left[ 2 \left( \alpha + \beta \right) \right]}{2} \)

= \( \Large \frac{1}{2}\{ 1-\frac{1-\tan^{2} \left( \alpha + \beta \right) }{1+\tan^{2} \left( \alpha + \beta \right) } \} \left[ \because \cos2 \alpha =\frac{1-\tan^{2} \alpha }{1+\tan^{2} \alpha } \right] \)

= \( \Large \frac{1}{2}\left[ 1 - \frac{1- \left(\frac{P}{1-q}\right)^{2} }{1 + \left(\frac{P}{1-q}\right)^{2} } \right] \)

=\( \Large \frac{1}{2}\left[ \frac{ \left(1-q\right)^{2}+P^{2}- \left(1-q\right)^{2}+P^{2} }{ \left(1-q\right)^{2}+P^{2} } \right] \)

= \( \Large \frac{P^{2}}{P^{2}+ \left(1-q\right)^{2} } \)