Question

Given that $\tan A,\tan B$ are the roots of the equation $x^2-px+q=0$, then the value of ${\sin }^2(A+B)$ is

Answer 1

From the given condition we have
$\tan A+\tan B=p,\tan A\tan B=q$
So that $\tan (A+B)=\frac{\tan A+tanB}{1-\tan A\tan B}=\frac{p}{1-q}$
And ${\sin }^2(A+B)=\frac{{\sin }^2(A+B)}{{\cos }^2(A+B)}\times {\cos }^2(A+B)=\frac{{\tan }^2(A+B)}{{\sec }^2(A+B)}=\frac{{\tan }^2(A+B)}{1+{\tan }^2(A+B)}=\frac{\left[\frac{p^2}{(1-q{)}^2}\right]}{[1+\frac{p^2}{(1-q{)}^2}]}=\frac{p^2}{p^2+(1-q{)}^2}$

​​​​​Hence the correct answer is Option A.