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Find the smallest positive value of $x+y$ if $(1+\tan x)(1+\tan y)=2$.

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$(1+\operatorname{tg} x)(1+\operatorname{tg} y)=2 \Leftrightarrow \operatorname{tg} x+\operatorname{tg} y=1-\operatorname{tg} x \operatorname{tg} y$.

Note that $1-\operatorname{tg} x \operatorname{tg} y \neq 0$. Indeed, if the sum of two numbers is zero, then they are of opposite signs, and their product is non-positive.

Therefore, the obtained equality can be written as $\operatorname{tg}(x+y)=1$. Hence, the smallest positive value of $x+y$ is $\pi / 4$.

## Answer