Trigonometry determine types of triangles?

From Sine Rule we know that $\displaystyle \frac{a}{b}=\frac{\sin A}{\sin B}$

Also, we are given that $\displaystyle \frac{a}{b}=\frac{\cos B}{\cos A}$

Equating the two equations give us

$\sin\left(2A\right)-\sin\left(2B\right)=0$

i.e.

$2\cos\left(A+B\right)\sin\left(A-B\right)=0$

Which implies that either $\displaystyle A=B$ meaning the triangle is isosceles or $\displaystyle A+B=\frac{\pi}{2}$ meaning that the third angle is a right angle and thus, the triangle is right angled.........

By the cosine rule, $a \cos A=b \cos B=a \Leftrightarrow a \cdot \frac{b^2+c^2-a^2}{2bc}=b \cdot \frac{a^2+c^2-b^2}{2ac}$ $\Leftrightarrow a^2 \left(b^2+c^2-a^2\right) = b^2 \left(a^2+c^2-b^2\right)$ $\Leftrightarrow a^2c^2-a^4-b^2c^2+b^4=0$ $\Leftrightarrow \left(a^2-b^2\right) \left(c^2-a^2-b^2\right)=0$ $\Leftrightarrow a=b \text{ or } a^2+b^2=c^2$ Thus, $\triangle ABC$ is isosceles or right triangle.