In a triangle $ABC$, if $a\ne \mathrm{b},\frac{a\cos A-b\cos B}{a\cos B-b\cos A}+\cos C=$

Given,

Triangle $ABC$,

To find the value of $\frac{a\cos A-b\cos B}{a\cos B-b\cos A}+\cos C$

Using the sine rule we get,

$=\frac{2R\sin A\cos A-2R\sin B\cos B}{2R\sin A\cos B-2R\sin B\cos A}+\cos C$

$=\frac{\sin 2A-\sin 2B}{2\left(\sin A\cos B-\sin B\cos A\right)}+\cos C$

$=\frac{2\sin \left(A-B\right)\cos \left(A+B\right)}{2\sin \left(A-B\right)}+\cos C$

$=\cos \left(A+B\right)+\cos C$

$=-\cos C+\cos C \left\{\text{as} A+B=\mathrm{\pi }-\mathrm{C}\right\}$ 

$=0$