3. Let $M=\{(x, y):|x y|=1, x>0\}, N=$ $\{(x, y): \operatorname{arctg} x+\operatorname{arcctg} y=\pi\}$. Then, ( )
(A) $M \cup N=\{(x, y):|x y|=1\}$
(B) $M \cup N=M$
(C) $M \cup N=N$
(D) $M \cup N=\{(x, y):|x y|=1$ and $x, y$ are not both negative\}
3. B.

In $M$, $|x y|=1$ is equivalent to $x y=1$ and $x y=-1$. But $x>0$, so it represents the graph of a hyperbolic function in the $\mathrm{I}$ and $\mathrm{N}$ quadrants.
In $N$, from $\operatorname{arctg} x+\operatorname{arcctg} y=\pi$, we get
$$
\operatorname{tg}(\operatorname{arctg} x+\operatorname{arcctg} y)=\frac{x+\frac{1}{y}}{1-\frac{x}{y}}=0 \text{. }
$$

Thus, $x y=-1$.
But from the given conditions and $00, y<0$.
Therefore, (B) holds.