If I am only given, in a triangle ABC, that:
$$\frac{a-c}{\cos B}= b(\cos C-\cos A)$$
$$\frac{a-c}{\cos C-\cos A}=\frac{b}{\cos B}$$
Would I be able to assume that
$$\frac{a}{\cos C}=\frac{c}{\cos A}=\frac{b}{\cos B} \,?$$
Since ratios like
$$\frac{x}{a}=\frac{y}{b}=k$$
are usually handled this way
$$\frac{x-y}{a-b}=\frac{x}{a}=\frac{y}{b}=k$$
In a question, I had to determine the nature of a triangle given the first equation at the very top, I successfully deduced that it was an isosceles triangle using the method mentioned above with the sine rule but I was told that it was incorrect. If so why?
no you can't.
for example, we have
$\frac{7-5}{9-5}=\frac{1}{2}$
but we don't have
$\frac{7}{9}=\frac{5}{5}=\frac{1}{2}$.