Question

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

• A. $\frac{a^2}{a^2+(1-b{)}^2}$
• B. $\frac{a^2}{a^2+b^2}$
• C. $\frac{a^2}{a^2-b^2}$
• D. $\frac{a^2}{b^2(1-a{)}^2}$

Answer 1

The correct option is A $\frac{a^2}{a^2+(1-b{)}^2}$

$x^2-ax+b=0$ has roots $\tan A,\tan B$
So, $\tan A+\tan B=a$
and $\tan A\cdot \tan B=b$
$\tan (A+B)=\frac{\tan A+\tan B}{1-\tan A\cdot \tan B}\Rightarrow \tan (A+B)=\frac{a}{1-b}$

Therefore, ${\sin }^2(A+B)$
$=\frac{1}{{\text{cosec}}^2(A+B)}=\frac{1}{1+{\cot }^2(A+B)}=\frac{1}{1+{\left(\frac{1-b}{a}\right)}^2}=\frac{a^2}{a^2+(1-b^2)}$