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

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### 3 Answers

We have to prove that tan A + tan B + tan C = tan A*tan B*tan C for any non-right angle triangle.

For any triangle the sum of the angles is equal to 180 degrees. If we take a triangle ABC, A + B + C = 180 degrees.

A + B + C = 180 or A + B = 180 - C

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

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

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

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

**This proves that tan A + tan B + tan C = tan A*tan B*tan C**

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.**

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