Question

lf $\tan A,\tan B$ are the roots of $x^2-4x+2=0$, then ${\cos }^2(A+B)$ is

• A. $\frac{4}{17}$
• B. $-\frac{4}{17}$
• C. $\frac{1}{17}$
• D. $-\frac{1}{17}$

Answer 1

The correct option is C $\frac{1}{17}$

$\tan A+\tan B=$ Sum of roots $=-(-4)=4$
$\tan A\tan B=$ Product of roots $=2$
$\tan (A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}=\frac{4}{1-2}=-4$
$\therefore \sin (A+B)=\frac{-4}{\sqrt{17}};\cos (A+B)=\frac{1}{\sqrt{17}}$
$\therefore {\cos }^2(A+B)=\frac{1}{17}$