$$
\begin{array}{l}
\text { 4. In } \triangle A B C, \\
\cos ^{2} A+\cos ^{2} B+\cos ^{2} C=\sin ^{2} B .
\end{array}
$$

Then $\tan A \cdot \tan C=$ $\qquad$
4. 1 or 3.

In $\triangle A B C$, $\cos B=-\cos (A+C)$.
Transforming the condition, we get
$$
\begin{aligned}
0= & \frac{1}{2}(1+\cos 2 A+1+\cos 2 C)+2 \cos ^{2} B-1 \\
= & \cos (A+C) \cdot \cos (A-C)+2 \cos ^{2}(A+C) \\
= & \cos (A+C) \cdot(\cos (A-C)+2 \cos (A+C)) \\
= & (\cos A \cdot \cos C-\sin A \cdot \sin C) \cdot \\
& (3 \cos A \cdot \cos C-\sin A \cdot \sin C) \\
= & \left.(\sin A \cdot \sin C)^{2}(\tan A \cdot \tan C 360)^{2} \cdot 73^{4}\right) \\
& (\tan A \cdot \tan C-3) \text { factor }
\end{aligned}
$$

Since $\sin A \cdot \sin C>0$, therefore,
$$
\tan A \cdot \tan C=1 \text { or } 3 \text {. }
$$