Question

If A+B+C = 180°, prove that: tan(A+B-C) + tan(B+C-A) + tan(C+A-B) = tan2A tan2b tan2C

Answer 1

Given that

A+B+C=180∘A+B+C=180∘

taking tan (tangent) on both the sides,

tan(A+B+C)=tan(180∘)tan⁡(A+B+C)=tan⁡(180∘)

tanA+tanB+tanC−tanAtanBtanC1−tanAtanB−tanBtanC−tanAtanC=0tan⁡A+tan⁡B+tan⁡C−tan⁡Atan⁡Btan⁡C1−tan⁡Atan⁡B−tan⁡Btan⁡C−tan⁡Atan⁡C=0

tanA+tanB+tanC−tanAtanBtanC=0tan⁡A+tan⁡B+tan⁡C−tan⁡Atan⁡Btan⁡C=0

tanA+tanB+tanC=tanAtanBtanCtan⁡A+tan⁡B+tan⁡C=tan⁡Atan⁡Btan⁡C

tanA+tanB+tanCtanAtanBtanC=1tan⁡A+tan⁡B+tan⁡Ctan⁡Atan⁡Btan⁡C=1