Show that if A, B, and C are the angles of an acute triangle, then tan A + tan B + tan C = tan A tan B tan C.

Question

I tried drawing perpendiculars and stuff but it doesn't seem to work?
For me, the trig identities don't seem to plug in as well.
Help is appreciated, thanks.

Answer 1

C = 180-(A+B), so
tan C = -tan(A+B) = (tanA+tanB)/(tanAtanB-1)

tanA+tanB+tanC
= tanA+tanB + (tanA+tanB)/(tanAtanB-1)
= (tanA+tanB)(1 + 1/(tanAtanB-1))
= (tanA+tanB)(tanAtanB)/(tanAtanB-1)
= (tanAtanB)(tanA+tanB)/(tanAtanB-1)
= tanAtanBtanC

Answer 2

Thanks a ton!