7. Let $M=\{(x, y):|x y|=1, x>0\}, N=\{(x, y): \arctan x+\arctan y=\pi\}$, 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 $\}$
(B)
7.【Analysis and Solution】In set $M$, $|xy|=1$ is equivalent to $xy=1$ or $xy=-1$ but $x>0$, so it represents the two branches of the hyperbolic function graph in the I and IV quadrants.
In set $N$, from $\arctan x + \operatorname{arccot} y = \pi$, we get
$$
\begin{aligned}
x & = \tan (\arctan x) = \tan (\pi - \operatorname{arccot} y) \\
& = -\tan (\operatorname{arccot} y) = -\frac{1}{\cot (\operatorname{arccot} y)} = -\frac{1}{y}.
\end{aligned}
$$

That is, $xy = -1$, which represents the two branches of the hyperbolic function curve in the II and IV quadrants.
Since $-\frac{\pi}{2} < \operatorname{arccot} y < \frac{\pi}{2}$, when $x > 0$, $0 < \operatorname{arccot} y < \frac{\pi}{2}$, and when $x < 0$, $-\frac{\pi}{2} < \operatorname{arccot} y < 0$. Therefore, $\{(x, y) \mid xy = -1, x > 0\}$ is the branch of the hyperbolic function curve $xy = -1$ in the IV quadrant, so $N \subset M$. Therefore, $M \cup N = M$.