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.

Sum of roots; i.e., $\tan A + \tan B = $ -coefficient of $x /$ coefficient of $x^2 = a$
Product of roots; i.e., $\tan A \tan B = $ constant $/$ coefficient of $x^2 = b$
Now, $\tan (A+B)= \dfrac{\tan A + \tan B}{1 - \tan A \tan B} = \dfrac{a}{1-b}$
Draw a right triangle with opp side as $a$ and adjacent side as $1-b$. 
So our hypotenuse will be $\sqrt{a^2 + (1-b)^2}$
$\sin(A+B)=$ opp/ hypotenuse $=\dfrac{a}{\sqrt{a^2 + (1-b)^2}}$
So $\sin^2(A+B)= \dfrac{a^2}{a^2 + (1-b)^2}$
therefore, option A :-)
