From - Fri Nov 28 11:43:48 1997
From: kubo@jacobi.math.brown.edu (Tal Kubo)
Subject: Re: Help with proving a certain identity??
Organization: Math Department, Brown University

> "For any acute or obtuse triangle,
> tanA + tanB + tanC = tanA * tanB * tanC"

For any angles,  tan(A+B+C)= (t1 - t3)/(t0-t2) where
tN is the N'th symmetric function of tan(A),tan(B),tan(C).
The identity above, t1=t3, is the numerator of tan(A+B+C)=0.
The denominator of tan(A/2+B/2+C/2)=infinity  yields the
analogous 

  1 = tan(A/2)tan(B/2)+tan(B/2)tan(C/2)+tan(C/2)tan(A/2).


                                t1 - t3 + t5 - t7 + ...
In general, tan(a+b+c+d+...) = -------------------------
                                t0 - t2 + t4 - t6 + ...