1. If $\frac{1+a}{1-a}=\frac{1-b}{1+b}$, then, $(2+a)(2+b)+b^{2}$
is equal to ( ).
(A) 4
(B) -4
(C) 2
(D) -2
1. A.

From the given, we have $(1+a)(1+b)=(1-a)(1-b)$, simplifying to $a+b=0$, which means $a=-b$.
Therefore, $(2+a)(2+b)+b^{2}=(2-b)(2+b)+b^{2}=4$.