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

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

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

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

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