If tan x

Given:

$\tan x=\frac{1}{2}$

$\tan y=\frac{1}{3}$

Since $\tan (x+y)=\frac{\tan x+\tan y}{1-\tan x \tan y}$

$=\frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{2} \times \frac{1}{3}}$

$=\frac{\frac{3+2}{6}}{\frac{6-1}{6}}$

$=\frac{5}{6} \times \frac{6}{5}$

$\tan (x+y)=1$

i. e. $x+y=\tan ^{-1}$

hence, value of $x+y$ is $\frac{\pi}{4}$