mathschallenge.net

Problem

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

Solution

As A + B + C = 180^o, it follows that tan(A + B + C) = tan180^o = 0.

Using the addition formula:

tan(A + (B + C))

=

tanA + tan(B + C) 1 tanA tan(B + C)

= 0

tanA + tan(B + C) = 0.

Using the addition formula again:

tan A +

tanB + tanC 1 tanB tanC

= 0

tanA(1 tanB tanC) + tanB + tanC = 0

tanA tanA tanB tanC + tanB + tanC = 0

Hence, tanA + tanB + tanC = tanA tanB tanC

As A + B = 180^o C, use tan(A + B) = tan(180^o C) to prove the same result by a slightly simpler route.