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:
$$\begin{align}
\text{R.H.S.}&=\dfrac {c-b}{a} \\[4pt]
&=\frac {a\cos B+b\cos A-a\cos C-c\cos A}{b\cos C+c\cos B} \\[4pt]
&=\dfrac {a(\cos B-\cos C)+(b-c)\cos A}{b\cos C+c\cos B}
\end{align}$$
 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.
A: Obvious if $B=C$
Otherwise  using this $$\dfrac{\cos B-\cos C}{\sin C-\sin B}=\cot\dfrac A2$$
$$\dfrac{1+\cos A}{\sin A}=?$$
A: For $\triangle ABC$,
$$\frac{\sin\left(A\right)}{a}=\frac{\sin\left(B\right)}{b}=\frac{\sin\left(C\right)}{c}=\frac{1}{k}$$
therefore,
$$k\sin\left(A\right)=a,\:k \sin\left(B\right)=b,\:k\sin\left(C\right)=c$$
\begin{align*}
RHS= &\frac{k\sin\left(C\right)-k\sin\left(B\right)}{k\sin\left(A\right)}\\&=\frac{\sin\left(C\right)-\sin\left(B\right)}{\sin\left(A\right)} \\&=\frac{2\cos\left(\frac{B+C}{2}\right)\cdot \sin\left(\frac{C-B}{2}\right)\cdot \cos\left(\frac{A}{2}\right)}{2\sin\left(\frac{A}{2}\right)\cdot \cos\left(\frac{A}{2}\right)\cdot \cos\left(\frac{A}{2}\right)}\\&=\frac{2\sin\left(\frac{A}{2}\right)\cdot \sin\left(\frac{C-B}{2}\right)\cdot \cos\left(\frac{A}{2}\right)}{2\sin\left(\frac{A}{2}\right)\cdot \cos^2\left(\frac{A}{2}\right)+1-1}\\&=\frac{2\sin\left(\frac{B+C}{2}\right)\cdot \sin\left(\frac{C-B}{2}\right)}{\cos\left(A\right)+1}\\&=\frac{\cos\left(B\right)-\cos\left(C\right)}{\cos\left(A\right)+1}
\end{align*}
