7. Let $M=\{(x, y) \mid | x y |=1, x>0\}, N=\{(x, y) \mid \arctan x+\operatorname{arccot} y=\pi\}$, 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 $\}$
When $x, y>0$, $\arctan x, \operatorname{arccot} y \in\left(0, \frac{\pi}{2}\right) ;$ when $x, y<0$, $\arctan x \in\left(-\frac{\pi}{2}, 0\right), \operatorname{arccot} y \in\left(0, \frac{\pi}{2}\right)$,
in all three cases, $\arctan x+\operatorname{arccot} y=\pi$ cannot be satisfied.
$x>0, y<0) \Rightarrow M \cup N=M$. Therefore, the answer is $B$.