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

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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...

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pialpha | Student

thanks!!!!!!

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giorgiana1976 | Student

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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