[Solution] Let $x=\operatorname{arctg} a$, $y=\operatorname{arctg} b$.
From
we have
$$
\begin{array}{c}
(a+1)(b+1)=2, \\
(\operatorname{tg} x+1)(\operatorname{tg} y+1)=2 .
\end{array}
$$

which means
$$
\operatorname{tg} x+\operatorname{tg} y=1-\operatorname{tg} x \operatorname{tg} y,
$$

or
$$
\operatorname{tg}(x+y)=\frac{\operatorname{tg} x+\operatorname{tg} y}{1-\operatorname{tg} x \operatorname{tg} y}=1 .
$$
Since $a=\frac{1}{2}>0$, we know $b>0$.
Thus $x, y \in\left[0, \frac{\pi}{2}\right]$, then $x+y \in[0, \pi]$.
Therefore,
$$
x+y=\frac{\pi}{4} \text{. }
$$

So the answer is $(C)$.
$7 \cdot 17$ If $a=\frac{1}{2},(a+1)(b+1)=2$, then the radian measure of $\operatorname{arctg} a+\operatorname{arctg} b$ is
(A) $\frac{\pi}{2}$.
(B) $\frac{\pi}{3}$.
(C) $\frac{\pi}{4}$.
(D) $\frac{\pi}{5}$.
(E) $\frac{\pi}{6}$.
(30th American High School Mathematics Examination, 1979)