<s>[INST] 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 . 
 [/INST] 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 $\alpha=\gamma$.
</s>