8. Let $\triangle A B C$ have internal angles $A, B, C$ with opposite sides $a, b, c$ respectively, and satisfy $a \cos B-b \cos A=\frac{4}{5} c$, then $\frac{\tan A}{\tan B}=$
$\qquad$ .
$$
\begin{array}{l}
\sin A \cos B-\sin B \cos A=\frac{4}{5} \sin C=\frac{4}{5}(\sin A \cos B+\sin B \cos A) \\
\Rightarrow \frac{1}{5} \sin A \cos B=\frac{9}{5} \sin B \cos A \Rightarrow \frac{\sin A \cos B}{\sin B \cos A}=9 \Rightarrow \frac{\tan A}{\tan B}=9
\end{array}
$$