Question

lf $\tan A,\tan B$ are the roots of $x^2-2x+2=0$, then ${\sin }^2(A+B)=$

• A. $\frac{4}{5}$
• B. $\frac{1}{2}$
• C. $\frac{3}{5}$
• D. $\frac{1}{4}$

Answer 1

The correct option is A $\frac{4}{5}$

$\tan A+\tan B=$ sum of roots $=\frac{-b}{a}=-\frac{-2}{1}=2$
$\tan A\tan B=\frac{c}{a}=\frac{2}{1}=2$
$\tan (A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}=\frac{2}{1-2}=-2.$
$\sin (A+B)=\frac{-2}{\sqrt{(2{)}^2+(1{)}^2}}=\frac{-2}{\sqrt{5}}$
$\therefore {\sin }^2(A+B)=\frac{4}{5}$