8. In $\triangle A B C$, $\cos B=\frac{1}{4}$. Then the minimum value of $\frac{1}{\tan A}+\frac{1}{\tan C}$ is $\qquad$ .
8. $\frac{2 \sqrt{15}}{5}$.

Given $\cos B=\frac{1}{4}$
$\Rightarrow \sin B=\sqrt{1-\cos ^{2} B}=\frac{\sqrt{15}}{4}$.
Then $\frac{1}{\tan A}+\frac{1}{\tan C}=\frac{\cos A}{\sin A}+\frac{\cos C}{\sin C}$
$$
=\frac{\sin B}{\sin A \cdot \sin C} \text {. }
$$

Notice that,
$$
\begin{array}{l}
0<\sin A \cdot \sin C \\
=-\frac{1}{2}(\cos (A+C)-\cos (A-C)) \\
=-\frac{1}{2}(-\cos B-\cos (A-C)) \\
=\frac{1}{8}+\frac{1}{2} \cos (A-C) \\
\leqslant \frac{1}{8}+\frac{1}{2}=\frac{5}{8},
\end{array}
$$

with equality holding if and only if $\angle A=\angle C$.
Thus, $\frac{1}{\tan A}+\frac{1}{\tan C} \geqslant \frac{\sqrt{15}}{4} \times \frac{8}{5}=\frac{2 \sqrt{15}}{5}$.
Therefore, the minimum value sought is $\frac{2 \sqrt{15}}{5}$.