Question 1161717
<pre>
We want to find:

{{{tan(A)tan(B)tan(C)}}}

we know 

{{{tan(A+B)= (tan(A)+tan(B))/(1-tan(A)*tan(B))}}}

Since (A+B+C=180^o}}} 

{{{A+B=180^o-C}}}

{{{tan(A+B)=tan(180^o-C) = -tan(C)}}}

so substitute in

{{{tan(A+B)= (tan(A)+tan(B))/(1-tanA*tanB)}}}

{{{-tan(C)= (tan(A)+tan(B))/(1-tan(A)*tan(B))}}} 

{{{-tan(C)+tan(A)*tan(B)*tan(C) = tan(A)+tan(B)}}}

{{{tan(A)*tan(B)*tan(C) = tan(A)+tan(B)+tan(C)=x}}}

So {{{tan(A)+tan(B)+tan(C)=x}}} also.

[It's interesting that the product and the sum of the tangents
of the angles of a non-right triangle are the same.  A right
triangle contains a 90° and tan(90°) is not defined.]

Edwin</pre>