Answer
$-\displaystyle \frac{\sqrt{2}}{2}$
Work Step by Step
Sketch two triangles,
one in which $t=\tan^{-1}2$,
(the hypotenuse is $c=\displaystyle \sqrt{4+1}=\sqrt{5}, \sin t=\frac{2}{\sqrt{}5}, \cos t=\frac{1}{\sqrt{}5})$,
and one in which $u=\tan^{-1}3$
(the hypotenuse is $d=\displaystyle \sqrt{9+1}=\sqrt{10}, \sin u=\frac{3}{\sqrt{10}}, \cos u=\frac{1}{\sqrt{10}}).$
Apply the additive identity for cosine:
$\cos(t+u)=\cos t\cos u-\sin t\sin u$
$=\displaystyle \frac{1}{\sqrt{5}}\cdot\frac{1}{\sqrt{10}}-\frac{2}{\sqrt{5}}\cdot\frac{3}{\sqrt{10}}=\frac{1-6}{\sqrt{50}}=\frac{-5}{5\sqrt{2}}=-\frac{\sqrt{2}}{2}$
