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tan-x-pi-4-3-tan-pi-9-tan-2pi-9-tan-x-pi-4-tan-pi-9-tan-2pi-9-

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Question Number 147643 by bobhans last updated on 22/Jul/21
tan (x+(π/4))+3(tan (π/9)+tan ((2π)/9))=tan (x+(π/4))tan (π/9)tan ((2π)/9)
tan(x+π4)+3(tanπ9+tan2π9)=tan(x+π4)tanπ9tan2π9
Answered by liberty last updated on 22/Jul/21
tan (x+(π/4))−tan (x+(π/4))tan (π/9)tan ((2π)/9)=−3(tan (π/9)+tan ((2π)/9)) tan (x+(π/4))[1−tan (π/9)tan ((2π)/9)]=−3(tan (π/9)+tan ((2π)/9)) tan (x+(π/4))=−3(((tan (π/9)+tan ((2π)/9))/(1−tan (π/9)tan ((2π)/9)))) tan (x+(π/4))=−3tan (π/3)=−3(√3) ⇒x+(π/4)=−arctan (3(√3))+nπ ⇒x=−(π/4)+nπ−arctan (3(√3))
tan(x+π4)−tan(x+π4)tanπ9tan2π9=−3(tanπ9+tan2π9)tan(x+π4)[1−tanπ9tan2π9]=−3(tanπ9+tan2π9)tan(x+π4)=−3(tanπ9+tan2π91−tanπ9tan2π9)tan(x+π4)=−3tanπ3=−33⇒x+π4=−arctan(33)+nπ⇒x=−π4+nπ−arctan(33)
Answered by gsk2684 last updated on 22/Jul/21
tan (x+(π/4))[1−tan (π/9)tan ((2π)/9)]=−3(tan (π/9)+tan ((2π)/9)) tan (x+(π/4))=−3(((tan (π/9)+tan ((2π)/9))/(1−tan (π/9)tan ((2π)/9)))) tan (x+(π/4))=−3 tan ((π/9)+((2π)/9)) ((tan x + 1)/(1− tan x)) =−3 tan (π/3) ((tan x + 1)/(1− tan x)) =−3(√3) tan x +1 = −3(√3) + 3(√3) tan x 3(√3)+1 = (3(√3) −1)tan x (3(√3)+1)^2 ={(3(√3))^2 −1} tan x 28+6(√(3 )) = 26 tan x ((14+3(√3))/(13))=tan x x=tan^(−1) ((14+3(√3))/(13))
tan(x+π4)[1−tanπ9tan2π9]=−3(tanπ9+tan2π9)tan(x+π4)=−3(tanπ9+tan2π91−tanπ9tan2π9)tan(x+π4)=−3tan(π9+2π9)tanx+11−tanx=−3tanπ3tanx+11−tanx=−33tanx+1=−33+33tanx33+1=(33−1)tanx(33+1)2={(33)2−1}tanx28+63=26tanx14+3313=tanxx=tan−114+3313

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