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
Solution: 4.
[Method 1] From the given and the cosine rule, we have
$$
a \cdot \frac{c^{2}+a^{2}-b^{2}}{2 c a}-b \cdot \frac{b^{2}+c^{2}-a^{2}}{2 b c}=\frac{3}{5} c \text {, i.e., } a^{2}-b^{2}=\frac{3}{5} c^{2} \text { . }
$$

Thus, $\frac{\tan A}{\tan B}=\frac{\sin A \cos B}{\sin B \cos A}=\frac{a \cdot \frac{c^{2}+a^{2}-b^{2}}{2 c a}}{b \cdot \frac{b^{2}+c^{2}-a^{2}}{2 b c}}=\frac{c^{2}+a^{2}-b^{2}}{b^{2}+c^{2}-a^{2}}=\frac{\frac{8}{5} c^{2}}{\frac{2}{5} c^{2}}=4$.
[Method 2] As shown in Figure 2, draw $C D \perp A B$, with the foot of the perpendicular at $D$, then
$$
a \cos B=D B, b \cos A=A D .
$$

From the given, we have $D B-A D=\frac{3}{5} c$.
Also, $D B+D A=c$.
Solving these equations simultaneously, we get $A D=\frac{1}{5} c, D B=\frac{4}{5} c$.
Thus, $\frac{\tan A}{\tan B}=\frac{\frac{C D}{A D}}{\frac{A D}{D B}}=\frac{D B}{A D}=4$.
[Method 3] From the projection theorem, we have $a \cos B+b \cos A=c$.
Also, $a \cos B-b \cos A=\frac{3}{5} c$.
Solving these equations simultaneously, we get $a \cos B=\frac{4}{5} c, b \cos A=\frac{1}{5} c$.
Thus, $\frac{\tan A}{\tan B}=\frac{\sin A \cos B}{\sin B \cos A}=\frac{a \cos B}{b \cos A}=\frac{\frac{4}{5} c}{\frac{1}{5} c}=4$.