In any $\triangle ABC$, prove that: $\frac {\cos B-\cos C}{\cos A +1}=\frac {c-b}{a}$

We have$$a\cos B+b\cos A=c,$$$$a\cos C+c\cos A=b.$$Subtracting gives$$a(\cos B-\cos C)-(c-b)\cos A=c-b,$$which on rearranging yields the required result.

Obvious if $B=C$

Otherwise using this $$\dfrac{\cos B-\cos C}{\sin C-\sin B}=\cot\dfrac A2$$