1. If $M=\left\{z \left\lvert\, z=\frac{t}{1+t}+i \frac{1+t}{t}\right., t \in \mathrm{R}, t \neq-1, t \neq 0\right\}, N=\{z \mid z$ $=\sqrt{2}[\cos (\arcsin t)+i \cos (\arccos t)], t \in \mathrm{R},|t| \leqslant 1\}$, then the number of elements in $M \cap N$ is
(A) 0
(B) 1
(C) 2
(D) 4
1. In the Cartesian coordinate system, the elements of sets $M$ and $N$ are the curves
$$
M: x y=1(x \neq 1, x \neq 0), N: x^{2}+y^{2}=2(0 \leqslant x \leqslant \sqrt{2}) \text {. }
$$

Thus, the x-coordinates of the intersection points of curves $M$ and $N$ should satisfy the equation $x^{2}+\frac{1}{x^{2}}=2$, solving which gives $x= \pm 1$, and $\pm 1 \notin(0,1) \cup(1, \sqrt{2})$. Therefore, $M \cap N=\varnothing$. Choose $A$.