Prove that tanA + tanB + tanC = tanA tanB tanC for any non-right angle triangle

The sum of the angles in any triangle is 180
degrees.

A+B+C = 180

A+B = 180
- C

We'll apply tangent
function:

tan (A+B) = tan (180 -
C)

We'll consider the
identity:

tan(x+y) = (tan x + tan y)/(1-tan x*tan
y)

(tan A + tan B)/(1-tan A*tan B) = (tan 180 - tan
C)/(1+tan 180*tan C)

But tan 180 = 0, therefore, we'll
get:

(tan A + tan B)/(1-tan A*tan B) = (0 - tan
C)/(1+0)

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

We'll multiply by (1-tan A*tan
B):

tan A + tan B = -tan C +tan A*tan B*tan
C

We'll add tan C:

tan A + tan
B+ tan C = tan A*tan B*tan C

We notice that
the given identity tan A + tan B+ tan C = tan A*tan B*tan C is
verified.