2. For the function $f(x)=\lg \frac{1+x}{1-x}$, there are three numbers satisfying $|a|<1,|b|<1,|c|<1$, and
$$
f\left(\frac{a+b}{1+a b}\right)=1, f\left(\frac{b-c}{1-b c}\right)=2 .
$$

Then, the value of $f\left(\frac{a+c}{1+a c}\right)$ is $(\quad)$.
(A) -1
(B) $\lg 2$
(C) $\sqrt{10}$
(D) 3
2.A.

Given $|a|<1,|b|<1,|c|<1$, we know
$$
\begin{array}{l}
\left|\frac{a+b}{1+a b}\right|<1,\left|\frac{b-c}{1-b c}\right|<1,\left|\frac{a+c}{1+a c}\right|<1 . \\
\text { and } f\left(\frac{a+b}{1+a b}\right)=\lg \frac{1+\frac{a+b}{1+a b}}{1-\frac{a+b}{1+a b}} \\
=\lg \frac{a b+1+a+b}{a b+1-a-b}=\lg \frac{(a+1)(b+1)}{(a-1)(b-1)}=1, \\
\text { hence } \frac{(a+1)(b+1)}{(a-1)(b-1)}=10 .
\end{array}
$$

Also, $f\left(\frac{b-c}{1-b c}\right)=\lg \frac{1+\frac{b-c}{1-b c}}{1-\frac{b-c}{1-b c}}$
$$
=\lg \frac{(1+b)(1-c)}{(1-b)(1+c)}=2 \text {, }
$$

hence $\frac{(1+b)(1-c)}{(1-b)(1+c)}=100$.
(1) and (2) give $\frac{(a+1)(c+1)}{(a-1)(c-1)}=\frac{1}{10}$.

Thus, $f\left(\frac{a+c}{1+a c}\right)=\lg \frac{(a+1)(c+1)}{(a-1)(c-1)}=\lg \frac{1}{10}=-1$.