Question:
If $\tan x=\frac{1}{2}$ and $\tan y=\frac{1}{3}$, then the value of $x+y$ is ________________
Solution:
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 $\frac{\pi}{4}$
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