<s>[INST] In any $\triangle ABC$, prove that: $\frac {\cos B-\cos C}{\cos A +1}=\frac {c-b}{a}$ In any $\triangle ABC$, prove that: $\dfrac {\cos B-\cos C}{\cos A +1}=\dfrac {c-b}{a}$
My Attempt:
$$\begin{align}
\text{R.H.S.}&=\dfrac {c-b}{a} \\[4pt]
&=\frac {a\cos B+b\cos A-a\cos C-c\cos A}{b\cos C+c\cos B} \\[4pt]
&=\dfrac {a(\cos B-\cos C)+(b-c)\cos A}{b\cos C+c\cos B}
\end{align}$$
 [/INST] 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.
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