How to prove the value of an trigonometric term to zero?

Question

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 .

Answer 1

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 $${\frac { \left( a-b \right) \left( a+b \right) }{c}}$$

Answer 2

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 $\alpha=\gamma$.