1. If $a, b$ are real numbers, satisfying $\frac{1+a}{1-a}=\frac{1-b}{1+b}$, then the value of $(1+a+b)(1-a-b)$ is $(\quad)$.
(A) -1
(B) 0
(C) 1
(D) 2
- 1. C.

From the given condition, we know that $1+a+b+ab=1-a-b+ab$ then $a+b=0$.