Chứng minh rằng trong tam giác ABC, ta có:

\(\tan \frac{{A + B - 2C}}{2} = \cot \frac{{3C}}{2}\).

Ta có: \(\frac{{A + B - 2C}}{2} = \frac{{A + B + C - 3C}}{2} = \frac{{\pi - 3C}}{2} = \frac{\pi }{2} - \frac{{3C}}{2}\).

Suy ra \(\tan \frac{{A + B - 2C}}{2} = \cot \frac{{3C}}{2}\).