Tangent identity given $a + b + c = \pi$

Question

Given that $a + b + c = \pi$, that is, three angles in a triangle - then prove that $$\tan a + \tan b + \tan c = \tan a \tan b \tan c$$

Is my solution below completely rigorous? Can I justify taking the tangent of both sides of my equation (I think not, since tangent isn't an injective function).

Answer 1

We can write $a + b = \pi -c$ then taking the tangent of both sides, this yields $$\tan (a +b) = \tan(\pi -c) \iff \frac{\tan a + \tan b}{1 - \tan a \tan b} = -\tan c$$

So $$\tan a + \tan b = \tan a \tan b \tan c - \tan c$$

Hence we arrive at $$\bbox[10px, border: blue 1px solid]{\tan a + \tan b + \tan c = \tan a \tan b \tan c} \quad \square$$

as required.

Answer 2

Note that; $\tan(a+b+c)=0$, then

\begin{eqnarray*} % \nonumber to remove numbering (before each equation) \frac{\tan(a+b)+\tan(c)}{1-\tan(a+b)\tan(c)} &=& 0 \\\\ \tan(a+b)+\tan(c) &=& 0 \\\\ \frac{\tan(a)+\tan(b)}{1-\tan(a)\tan(b)}+\tan(c)&=& 0 \\\\ \tan(a)+\tan(b)+(1-\tan(a)\tan(b))\tan(c) &=& 0 \\\\ \tan(a)+\tan(b)+\tan(c)-\tan(a)\tan(b)\tan(c) &=& 0\\\\ \tan(a)+\tan(b)+\tan(c) &=& \tan(a)\tan(b)\tan(c) \end{eqnarray*}