Ta có:
$\quad \tan(x+y+z)=\tan[(x+y) + z]$
$\to \tan(x+y+z)=\dfrac{\tan(x+y)+\tan z}{1 - \tan(x+y)\tan z}$
$\to \tan(x+y+z)=\dfrac{\dfrac{\tan x + \tan y}{1 -\tan x.\tan y} + \tan z}{1 - \dfrac{\tan x +\tan y}{1-\tan x.\tan y}\cdot\tan z}$
$\to \tan(x+y+z)=\dfrac{\tan x + \tan y +\tan z - \tan x.\tan y.\tan z}{1- \tan x.\tan y-\tan y.\tan z - \tan z.\tan x}$
Đặt $\begin{cases}x = A - B\\y = B - C\\z = C - A\end{cases}$
Ta được:
$\quad\tan(x + y + z) = \tan(A-B + B - C + C -A)$
$\to \tan(x+y+z)= \tan0$
$\to \tan(x+y+z)= 0$
$\to \dfrac{\tan x + \tan y +\tan z - \tan x.\tan y.\tan z}{1- \tan x.\tan y-\tan y.\tan z - \tan z.\tan x} = 0$
$\to \tan x + \tan y +\tan z - \tan x.\tan y.\tan z = 0$
$\to \tan x + \tan y +\tan z = \tan x.\tan y.\tan z$
Hay $\tan(A-B) + \tan(B-C) + \tan(C-A) = \tan(A-B).\tan(B-C).\tan(C-A)$
