In $\Delta ABC$ prove that $\frac{a-b\cos C}{b-a\cos C}=\frac{\cos B}{\cos A}$

Question

LHS:

$$\frac{a-b(\frac{a^2+b^2-c^2}{2ab})}{b-a(\frac{a^2+b^2-c^2}{2ab})}$$ $$\frac{a^2+c^2-b^2}{b^2+c^2-a^2}$$

I can’t solve further. You can’t devide the numerator and denominate by $2ac$ or $2bc$, otherwise it would have been really easy

Answer 1

Continue with

$$\frac{a-b(\frac{a^2+b^2-c^2}{2ab})}{b-a(\frac{a^2+b^2-c^2}{2ab})}= \frac ba \cdot \frac{a^2+c^2-b^2}{b^2+c^2-a^2} =\frac{\frac{a^2+c^2-b^2}{2ac}}{\frac{b^2+c^2-a^2}{2bc}}=\frac{\cos B}{\cos A}$$