How to prove the value of an trigonometric term to zero? Given that , in $\triangle ABC$ , $AC \neq BC$ . We have to prove that , $\dfrac{BC\cos C-AC\cos B}{BC\cos B-AC\cos A}+\cos C=0$
My trying:
As $AC \neq BC \Rightarrow b \neq a \Rightarrow a=c \Rightarrow A=C$ . So for the following term , $$ \\ $$
\begin{align}
\frac{BC\cos C-AC\cos B}{BC\cos B-AC\cos A}+\cos C
&= \frac{a\cos C-b\cos B}{a\cos B-b\cos A}+\cos C = \frac{a\cos A-b\cos B}{a\cos B-b\cos A}+\cos A \\[1ex]
&= \frac{a\cos A-b\cos B+a\cos A\cos B=b{\cos }^2A}{a\cos B-b\cos A}
\end{align}
Then I got stuck and have no clue how to go ahead . Can anyone please help me to solve this problem ? It will be of great help . 
 A: for the first numerator i have got $$1/2\,{\frac {{a}^{2}+{b}^{2}-{c}^{2}}{b}}-1/2\,{\frac {b \left( {a}^{2
}-{b}^{2}+{c}^{2} \right) }{ac}}
$$
for the denominator i have got
$$1/2\,{\frac {{a}^{2}-{b}^{2}+{c}^{2}}{c}}-1/2\,{\frac {-{a}^{2}+{b}^{2
}+{c}^{2}}{c}}
$$
and the sum is given by
$${\left( 1/2\,{\frac {{a}^{2}+{b}^{2}-{c}^{2}}{b}}-1/2\,{\frac {b
 \left( {a}^{2}-{b}^{2}+{c}^{2} \right) }{ac}} \right)  \left( 1/2\,{
\frac {{a}^{2}-{b}^{2}+{c}^{2}}{c}}-1/2\,{\frac {-{a}^{2}+{b}^{2}+{c}^
{2}}{c}} \right) ^{-1}}+1/2\,{\frac {{a}^{2}+{b}^{2}-{c}^{2}}{ab}}
$$
have you got this?
the simplified numerator is given by $$1/2\,{\frac {{a}^{3}c-{a}^{2}{b}^{2}+a{b}^{2}c-a{c}^{3}+{b}^{4}-{b}^{2
}{c}^{2}}{bac}}
$$
and the simplified denominator is given by $${\frac { \left( a-b \right)  \left( a+b \right) }{c}}$$
A: Let's write $BC=a$, $AC=b$ and $AB=c$, and the opposite angles $\widehat{BAC}=\alpha$, $\widehat{ABC}=\beta$, $\widehat{ACB}=\gamma$, so your equation becomes
$$
\frac{a\cos\gamma-b\cos\beta}{a\cos\beta-b\cos\alpha}+\cos\gamma=0
$$
The sine law tells us that, if $R$ is the circumradius,
$$
a=2R\sin\alpha
\qquad
b=2R\sin\beta
$$
so we can rewrite the equation as
$$
\frac{\sin\alpha\cos\gamma-\sin\beta\cos\beta}{\sin\alpha\cos\beta-\sin\beta\cos\alpha}+\cos\gamma=0
$$
or
$$
\sin\alpha\cos\gamma(1+\cos\beta)-\sin\beta(\cos\beta+\cos\alpha\cos\gamma)=0
$$
Since $\beta=\pi-(\alpha+\gamma)$, we easily see that $\cos\beta+\cos\alpha\cos\gamma=\sin\alpha\sin\gamma$ and the equation becomes
$$
\sin\alpha\cos\gamma(1+\cos\beta)-\sin\alpha\sin\beta\sin\gamma=0
$$
or (as $\sin\alpha\ne0$)
$$
\cos\gamma+\cos\beta\cos\gamma-\sin\beta\sin\gamma=0
$$
that is, $\cos\gamma+\cos(\beta+\gamma)=0$ and therefore $\cos\gamma=\cos\alpha$.
So the stated equality implies the angles at $A$ and $C$ are equal. Your triangle must be isosceles on the base $AC$.
Tracing the steps back, we set that the given equality holds for $\alpha=\gamma$.
