trigonometrical inequality $\tan{A} + \tan{B} + \tan{C} = \tan{A}\tan{B}\tan{C}$, Hint to prove $S2\neq 1$? I was studying conditional identities for triangle in trigonometry where I had to prove that
$\tan{A} + \tan{B} + \tan{C} = \tan{A}\tan{B}\tan{C}$
So I started with $\tan({A+B+C)}=\frac{S1-S3}{1-S2}$ where $S1=\sum_{cyc}\tan A$,  $S2=\sum_{cyc}\tan A.\tan B$ and  $S3=\prod_{cyc}\tan A$
Now since $A+B+C=\pi$ and since $\tan\pi=0$, therefore $\frac{S1-S3}{1-S2}=0$ and hence $S1=S3$ but I got stuck when I realized that another condition, $1-S2\neq0$ should also be true. Now I can't prove this.
I'm not having any idea how to start to prove $S2\neq1$ for a triangle. please give me a hint.

$~~~~\tan A\tan B+\tan B\tan C+\tan A\tan C\\=\tan A \cdot [\tan (B+C) \cdot (1-\tan B ~tan C)]+\tan B\tan C\\
=-(\tan A)^2 \cdot  (1-\tan B ~tan C)+\tan B\tan C\\
=(\tan^2 A+1)(\tan B \tan C)-\tan^2 A$
Assume there exist solution for $~\tan A\tan B+\tan B\tan C+\tan A\tan C=1$
$⇒ (\tan^2 A+1)(\tan B \tan C)=\tan^2 A+1$
$⇒\tan B \tan C=1$
$~~~~~\tan A=-\tan (B+C)=\frac{\tan B+\tan C}{1-\tan B \tan C}~~$ which is not defined $($as denominator $= 0)$.
Hence $~\tan A\tan B+\tan B\tan C+\tan A\tan C \ne 1$
