Inverse trigonometry - how does this work?

I'm learning inverse trigonometry functions from reference books, and certain questions bother me... They're of the type - Prove that $$\arctan(a) + \arctan(b) + \arctan(c) = \pi$$

And they're usually done by adding the angles in terms of $\tan$, but if $\tan(f) = 0$, $f$ should be $\pi$ or $0$, so how does it serve as a concrete proof?

EDIT

The proof usually follows the following "steps" - $$let arctan(a) = \alpha, arctan(b) = \beta, arctan(c) = \gamma$$ $$Also, tan(\alpha + \beta + \gamma) = \frac{tan(\alpha) + tan(\beta) + tan(\gamma) - tan(\alpha)tan(\beta)tan(\gamma)}{1 - tan(\alpha)tan(\beta) - tan(\beta)tan(\gamma) - tan(\gamma)tan(\alpha)}$$

And then we subsitute the values into the second identity, and the answer comes out to be zero, but that doesn't quite prove that the sum of the angles is $\pi$, it could also be $0$, which is my question

• Do you care to give a complete example of a proof your question is about? – Moritz Sep 22 '16 at 9:09
• I'm editing it right now, since I'm new to mathjax... – Shahe Ansar Sep 22 '16 at 9:23
• Is that okay, or should I elaborate it more? – Shahe Ansar Sep 22 '16 at 9:30
• @Jean-ClaudeArbaut Oh, makes sense, thanks – Shahe Ansar Sep 22 '16 at 9:36
• @YvesDaoust You misunderstood my question, we need to prove that the sum of arc-tans of three numbers equals pi, and those numbers are given in the question, you don't arbitrarily choose them. Since they ARE equal to pi, the tangent of their sum must be zero, but t hat's true when their sum is 0 and pi, so how do we know that it is pi, was my question – Shahe Ansar Sep 22 '16 at 9:39

Knowing only that $\tan x=0$ only tells you that $x=k\pi$.
In your case, if you know that $a,b,c$ are positive, then their arctangent is in $]0,\pi/2[$, and you can go further: the sum of these arctangents has to be in $]0,3\pi/2[$, and if the tangent of this sum is $0$, then the sum is $\pi$.