Prove tanA+tanB+tanC=tanAtanBtanC given that A+B+C=180 deg?

Question

Thanks to kataoun and railrule. katsaoun gets the nod for brevity, as soon as i find how to navigate to the appropriate icon. I may be able to learn from railrules's answer also.

Answer 1

You have
A+B=180-C
Tan(A+B)=Tan(180-C)
[tanA+tanB]/[1-tanAtanB]=-tanC
tanA+tanB=-tanC+tanAtanBtanC
tanA+tanB+tanC=tanAtanBtanC

Answer 2

hola

C = 180* - (A+B)

tan A + tan B =
= (sin A/ cos A) + (sin B/ cos B)
= ( sin A cos B + cos A sin B)/ (cos A cos B)
= sin (A+B) / (cos A cos B)
= sin C / (cos A cos B)

tan A + tan B + tan C =
= ( sin C / (cos A cos B)) + (sin C / cos C)
= sin C * ( cos C + cos A cos B )/ (cos A cos B cos C )

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cos C = cos ( 180* - (A+B))
= - cos (A+B)
= - cos A cos B + sin A sin B
cos C + cos A cos B = sin A sin B

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= sin C * (sin A sin B ) / (cos A cos B cos C )
= tan A * tan B * tan C


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