7・21 Let $M=\{(x, y) \mid | x y | = 1, x > 0\}$,
$$
N=\{(x, y) \mid \operatorname{arctg} x + \operatorname{arcctg} y = \pi\},
$$

then
(A) $M \cup N=\{(x, y) \mid | x y | = 1\}$.
(B) $M \cup N=M$.
(C) $M \cup N=N$.
(D) $M \cup N=\{(x, y) \mid | x y | = 1$ and $x, y$ are not both negative $\}$. (China High School Mathematics League, 1982)
[Solution] In $M$, $|x y|=1$ is equivalent to $x y=1$ and $x y=-1$, but $x>0$, so it represents the branches of the hyperbola in the I and IV quadrants.
In $N$, from $\operatorname{arctg} x+\operatorname{arcctg} y=\pi$, we get

Then
$$
\begin{array}{c}
\operatorname{tg}(\operatorname{arctg} x+\operatorname{arcctg} y)=\frac{x+\frac{1}{y}}{1-\frac{x}{y}}=0, \\
x y=-1 .
\end{array}
$$

But from the given condition and $00 \quad(\because x y=-1)$, thus
$$
00$, therefore $N \subset M, M \cup N=M$. Hence, the answer is $(B)$.