1. If $a>1, b>1$, and $\lg (a+b)=\lg a+$ $\lg b$, then the value of $\lg (a-1)+\lg (b-1)$ is ( ).
(A) equal to $\lg 2$
(B) equal to 1
(C) equal to 0
(D) not a constant independent of $a$ and $b$
$$
\begin{array}{l}
-1 . C . \\
\because \lg (a+b)=\lg a+\lg b, \\
\therefore a+b=a b, \text { i.e., }(a-1)(b-1)=1 .
\end{array}
$$

Therefore, $\lg (a-1)+\lg (b-1)=0$.