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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thanks!!!!!!
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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