2. Let $\triangle A B C$ have internal angles $A, B, C$ with opposite sides $a, b, c$ respectively, and satisfy $a \cos B - b \cos A = \frac{3}{5} c$. Then the value of $\frac{\tan A}{\tan B}$ is . $\qquad$
$$
\begin{array}{l}
a \cos B - b \cos A = \frac{3}{5} c \Rightarrow \sin A \cos B - \sin B \cos A = \frac{3}{5}(\sin A \cos B + \sin B \cos A) \\
\Rightarrow 2 \sin A \cos B = 8 \sin B \cos A \Rightarrow \tan A = 4 \tan B \Rightarrow \frac{\tan A}{\tan B} = 4
\end{array}
$$