1. If $a>1, b>1$, and $\lg (a+b)=\lg a+\lg b$, then the value of $\lg (a-1)+\lg (b-1)$
A. equals $\lg 2$
B. equals 1
C. equals 0
D. is not a constant independent of $a$ and $b$
$\lg (a+b)=\lg a+\lg b \Rightarrow a+b=a b$, thus $(a-1)(b-1)=a b-(a+b)+1=1$ $\Rightarrow \lg (a-1)+\lg (b-1)=0$. The derived conclusion is.

The translation is as follows:

$\lg (a+b)=\lg a+\lg b \Rightarrow a+b=a b$, thus $(a-1)(b-1)=a b-(a+b)+1=1$ $\Rightarrow \lg (a-1)+\lg (b-1)=0$. The derived conclusion is.