134 Given $a_{k}=\tan k \cdot \tan (k-1), k=1,2, \cdots, a_{1}+a_{2}+\cdots+a_{1994}=$
$134 \frac{\tan 1994}{\tan 1}-1994$. Applying the formula
$$
\tan 1=\frac{\tan k-\tan (k-1)}{1+\tan k \tan (k-1)}
$$

we get
$$
\begin{aligned}
a_{1}+a_{2}+\cdots+a_{1994} & =\sum_{k=1}^{1994} \tan k \tan (k-1)=\sum_{k=1}^{1994}\left(\frac{\tan k-\tan (k-1)}{\tan 1}-1\right) \\
& =\sum_{k=1}^{1994} \frac{\tan k}{\tan 1}-\sum_{k=1}^{1993} \frac{\tan k}{\tan 1}-1994 \\
& =\frac{\tan 1994}{\tan 1}-1994 .
\end{aligned}
$$