Originally Posted by
mathaddict True since this comes under trigonometry, my teacher wouldn't accept it either. WTell, i will try another way.
$\displaystyle \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$ from the sine rule.
Lets begin from the LHS, divide both the denominator and numerator by a,
$\displaystyle \frac{\cos A-\frac{b}{a}\cos B}{\frac{b}{a}\cos A-\cos B}$
$\displaystyle = \frac{\cos A- \frac{\sin B}{\sin A}\cdot \cos B}{\frac{\sin B}{\sin A}\cdot \cos A-\cos B}$
$\displaystyle =\frac{\sin 2A-\sin 2B}{2\sin (B-A)}$
$\displaystyle =\frac{\cos (A+B)\sin (A-B)}{-\sin (A-B)}$
$\displaystyle =-\cos (A+B)$
$\displaystyle =-\cos (180-B+B-C)$
$\displaystyle =\cos C$
Hope this works.