Finding (sin(A+B))^2 given roots of a quadratic equation. If tan A and tan B are the roots of the equation x^2 -ax + b = 0, then the value of sin(A+B)^2 is?
Options are:
((a^2)/((a^2)+(1-b)^2),
(a^2)/(a^2+b^2), 
a^2/(b+a)^2,
a^2/(b^2*(1-a)^2)
The value is needed in terms of a and b. 
The farthest i have gotten is the sum of roots giving me:
(sinAcosB + CosAsinB)/CosAcosB = a
which simplifies to sin^2(A+B) = (a*cosA*cosB)^2
after which i substituted the value obtained from product of roots which was ((sinAsinB)/b)^2 = (CosACosB)^2 which is substituted again to obtain:
(a*SinA*sinB)^2/(b^2)
Please find the value with steps as i am very confused about this question.

So, $\displaystyle\tan A+\tan B=\frac a1$ and $\displaystyle \tan A\tan B=\frac b1$
$\displaystyle\tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}$
$\displaystyle\sin^2(A+B)=\frac1{\csc^2(A+B)}=\frac1{1+\cot^2(A+B)}=\frac{\tan^2(A+B)}{\tan^2(A+B)+1}=\cdots$
