7. 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 $\}$.
7. (B)

In the set $M$, $|x y|=1$ is equivalent to $x y=1$ or $x y=-1$ but $x>0$, hence it represents the branches of the hyperbolic function graph in the I and IV quadrants.
In the set $N$, from $\operatorname{arctg} x+\operatorname{arcctg} y=\pi$ we get
$$
\begin{aligned}
x & =\operatorname{tg}(\operatorname{arctg} x)=\operatorname{tg}(\pi-\operatorname{arcctg} y) \\
& =-\operatorname{tg}(\operatorname{arcctg} y)=-\frac{1}{\operatorname{ctg}(\operatorname{arcctg} y)}=-\frac{1}{y} .
\end{aligned}
$$

That is, $x y=-1$ which represents the two branches of the hyperbolic function curve in the II and III quadrants.

Since $-\frac{\pi}{2}<\operatorname{arctg} x<\frac{\pi}{2}$ and $0<\operatorname{arcctg} y<\pi$, when $x>0$, $0<\operatorname{arcctg} y<\frac{\pi}{2}$, and when $x<0$, $\frac{\pi}{2}<\operatorname{arcctg} y<\pi$.
Thus, the set $\{ (x, y) \mid x y=-1, x>0 \}$ is the branch of the hyperbolic function curve $x y=-1$ in the IV quadrant, and the set $\{ (x, y) \mid x y=-1, x<0 \}$ is the branch of the hyperbolic function curve $x y=-1$ in the II quadrant.
It is known that $N \subset M \quad \therefore M \cup N=M$.