Trigonometric Property when $a+b+c=abc$

Question

This isn't the actual problem but I saw an answer where someone used this property to solve the problem . It says that if we have $$a+b+c=abc$$ then $\exists \alpha,\beta,\gamma \:$ $a=tan(\alpha)\:$;$b=tan(\beta)\:$;$c=tan(\gamma)\:$ where $\alpha+\beta+\gamma=\pi$
I don't know how to prove it , I hope someone gonna help me out !
By the way $a,b,c$ are positive reals

Answer 1

Let $a=\tan\alpha$, $b=\tan\beta$ and $c=\tan\gamma$.

Thus, $-\frac{3\pi}{2}<\alpha+\beta+\gamma<\frac{3\pi}{2}.$

We have$$\tan\alpha+\tan\beta+\tan\gamma(1-\tan\alpha\tan\beta)=0$$ or $$\tan(\alpha+\beta)+\tan\gamma=0$$ or $$\frac{\sin(\alpha+\beta+\gamma)}{\cos(\alpha+\beta)\cos\gamma}=0,$$ which gives $\alpha+\beta+\gamma=\pi k$, where $k\in\mathbb Z$.

If $\alpha+\beta+\gamma=\pi$ then we are done.

If $\alpha+\beta+\gamma=0$ then $\alpha+\beta+(\gamma+\pi)=\pi.$

If $\alpha+\beta+\gamma=-\pi$ then $\alpha+\beta+(\gamma+2\pi)=\pi.$

Answer 2

We have that $$\tan (x+y+z)=\frac {\tan x +\tan y+\tan z-\tan x\tan y\tan z}{1-\tan x\tan y-\tan y\tan z-\tan z\tan x}$$

Now for positive real numbers $a, b, c$ we can find $x, y, z \in (0, \frac {\pi} 2)$ with $x=\arctan a, y=\arctan b, z= \arctan c$ so that $\tan (x+y+z)=0$.

We have $x+y+z \in (0,\frac {3\pi}2)$ and the only possible value is $\pi$