Question

If $\tan A$ and $\tan B$ are the roots of the quadratic equation $x^2-ax+b=0,$ then ${\sin }^2(A+B)$ is

Answer 1

We want to find ${\sin }^2(A+B)$.
If we know any basic trigonometric ratio of the angle A+B, we can find ${\sin }^2(A+B)$ easily.
Since $\tan A$ and $\tan B$ are the roots of given quadratic equation, we can find $\tan A+\tan B$ and $\tan A\tan B$.
Once we have these two, we can find $\tan (A+B)$.

$\tan A+\tan B=a$

$\tan A\tan B=b$

$\Rightarrow \tan (A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}=\frac{a}{1-b}$

$\Rightarrow \sin (A+B)=\pm \frac{a}{\sqrt{a^2+(1-b{)}^2}}$
$\Rightarrow {\sin }^2(A+B)=\frac{a^2}{a^2+(1-b{)}^2}$