Finding (sin(A+B))^2 given roots of a quadratic equation.

Question

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.

Answer 1

So, $\displaystyle\tan A+\tan B=\frac a1$ and $\displaystyle \tan A\tan B=\frac b1$

$\displaystyle\tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}$

$\displaystyle\sin^2(A+B)=\frac1{\csc^2(A+B)}=\frac1{1+\cot^2(A+B)}=\frac{\tan^2(A+B)}{\tan^2(A+B)+1}=\cdots$

Answer 2

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 :-)