Trigonometric Property when a+b+c=abcthen ∃α,β,γ a=tan⁡α; b=tan⁡β;c=tan⁡γ where α+β+γ=πI don't know how to prove it

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Trigonometric Property when a+b+c=abcthen ∃α,β,γ a=tan⁡α; b=tan⁡β;c=tan⁡γ where α+β+γ=πI don&#039;t know how to prove it

Wrastirtyzp9w · Accepted Answer

Let a=tan⁡α, b=tan⁡β and c=tan⁡γ.Thus, −3π2&amp;lt;α+β+γ&amp;lt;3π2We havetan⁡α+tan⁡β+tan⁡γ(1−tan⁡αtan⁡β)=0ortan⁡(α+β)+tan⁡γ=0orsin⁡(α+β+γ)cos⁡(α+β)cos⁡γ=0which gives α+β+γ=πk, where k∈ZIf α+β+γ=π then we are done.If α+β+γ=0 then α+β+(γ+π)=πIf α+β+γ=−π then α+β+(γ+2π)=π

fonne0kgq · Accepted Answer

We have that tan⁡(x+y+z)=tan⁡x+tan⁡y+tan⁡z−tan⁡xtan⁡ytan⁡z1−tan⁡xtan⁡y−tan⁡ytan⁡z−tan⁡ztan⁡xNow for positive real numbers a,b,c we can find x,y,z∈(0,π2) withx=arctan⁡a,y=arctan⁡b,z=arctan⁡c so that tan⁡(x+y+z)=0We have x+y+z∈(0,3π2) and the only possible value is π

Trigonometric Property when a+b+c=abc then \(\displaystyle\exists\alpha,\beta,\gamma\) \(\displaystyle{a}={\tan{\alpha}};\

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