Question Number 147643 by bobhans last updated on 22/Jul/21
$$\:\mathrm{tan}\:\left(\mathrm{x}+\frac{\pi}{\mathrm{4}}\right)+\mathrm{3}\left(\mathrm{tan}\:\frac{\pi}{\mathrm{9}}+\mathrm{tan}\:\frac{\mathrm{2}\pi}{\mathrm{9}}\right)=\mathrm{tan}\:\left(\mathrm{x}+\frac{\pi}{\mathrm{4}}\right)\mathrm{tan}\:\frac{\pi}{\mathrm{9}}\mathrm{tan}\:\frac{\mathrm{2}\pi}{\mathrm{9}} \\ $$
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))](https://www.tinkutara.com/question/Q147648.png)
$$\mathrm{tan}\:\left({x}+\frac{\pi}{\mathrm{4}}\right)−\mathrm{tan}\:\left({x}+\frac{\pi}{\mathrm{4}}\right)\mathrm{tan}\:\frac{\pi}{\mathrm{9}}\mathrm{tan}\:\frac{\mathrm{2}\pi}{\mathrm{9}}=−\mathrm{3}\left(\mathrm{tan}\:\frac{\pi}{\mathrm{9}}+\mathrm{tan}\:\frac{\mathrm{2}\pi}{\mathrm{9}}\right) \\ $$$$\mathrm{tan}\:\left({x}+\frac{\pi}{\mathrm{4}}\right)\left[\mathrm{1}−\mathrm{tan}\:\frac{\pi}{\mathrm{9}}\mathrm{tan}\:\frac{\mathrm{2}\pi}{\mathrm{9}}\right]=−\mathrm{3}\left(\mathrm{tan}\:\frac{\pi}{\mathrm{9}}+\mathrm{tan}\:\frac{\mathrm{2}\pi}{\mathrm{9}}\right) \\ $$$$\mathrm{tan}\:\left({x}+\frac{\pi}{\mathrm{4}}\right)=−\mathrm{3}\left(\frac{\mathrm{tan}\:\frac{\pi}{\mathrm{9}}+\mathrm{tan}\:\frac{\mathrm{2}\pi}{\mathrm{9}}}{\mathrm{1}−\mathrm{tan}\:\frac{\pi}{\mathrm{9}}\mathrm{tan}\:\frac{\mathrm{2}\pi}{\mathrm{9}}}\right) \\ $$$$\mathrm{tan}\:\left({x}+\frac{\pi}{\mathrm{4}}\right)=−\mathrm{3tan}\:\frac{\pi}{\mathrm{3}}=−\mathrm{3}\sqrt{\mathrm{3}} \\ $$$$\Rightarrow{x}+\frac{\pi}{\mathrm{4}}=−\mathrm{arctan}\:\left(\mathrm{3}\sqrt{\mathrm{3}}\right)+{n}\pi \\ $$$$\Rightarrow{x}=−\frac{\pi}{\mathrm{4}}+{n}\pi−\mathrm{arctan}\:\left(\mathrm{3}\sqrt{\mathrm{3}}\right) \\ $$
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))](https://www.tinkutara.com/question/Q147649.png)
$$\mathrm{tan}\:\left({x}+\frac{\pi}{\mathrm{4}}\right)\left[\mathrm{1}−\mathrm{tan}\:\frac{\pi}{\mathrm{9}}\mathrm{tan}\:\frac{\mathrm{2}\pi}{\mathrm{9}}\right]=−\mathrm{3}\left(\mathrm{tan}\:\frac{\pi}{\mathrm{9}}+\mathrm{tan}\:\frac{\mathrm{2}\pi}{\mathrm{9}}\right) \\ $$$$\mathrm{tan}\:\left({x}+\frac{\pi}{\mathrm{4}}\right)=−\mathrm{3}\left(\frac{\mathrm{tan}\:\frac{\pi}{\mathrm{9}}+\mathrm{tan}\:\frac{\mathrm{2}\pi}{\mathrm{9}}}{\mathrm{1}−\mathrm{tan}\:\frac{\pi}{\mathrm{9}}\mathrm{tan}\:\frac{\mathrm{2}\pi}{\mathrm{9}}}\right) \\ $$$$\mathrm{tan}\:\left({x}+\frac{\pi}{\mathrm{4}}\right)=−\mathrm{3}\:\mathrm{tan}\:\left(\frac{\pi}{\mathrm{9}}+\frac{\mathrm{2}\pi}{\mathrm{9}}\right) \\ $$$$\frac{\mathrm{tan}\:{x}\:+\:\mathrm{1}}{\mathrm{1}−\:\mathrm{tan}\:{x}}\:=−\mathrm{3}\:\mathrm{tan}\:\frac{\pi}{\mathrm{3}} \\ $$$$\frac{\mathrm{tan}\:{x}\:+\:\mathrm{1}}{\mathrm{1}−\:\mathrm{tan}\:{x}}\:=−\mathrm{3}\sqrt{\mathrm{3}} \\ $$$$\mathrm{tan}\:{x}\:+\mathrm{1}\:\:=\:−\mathrm{3}\sqrt{\mathrm{3}}\:+\:\mathrm{3}\sqrt{\mathrm{3}}\:\mathrm{tan}\:{x} \\ $$$$\mathrm{3}\sqrt{\mathrm{3}}+\mathrm{1}\:=\:\left(\mathrm{3}\sqrt{\mathrm{3}}\:−\mathrm{1}\right)\mathrm{tan}\:{x} \\ $$$$\left(\mathrm{3}\sqrt{\mathrm{3}}+\mathrm{1}\right)^{\mathrm{2}} =\left\{\left(\mathrm{3}\sqrt{\mathrm{3}}\right)^{\mathrm{2}} −\mathrm{1}\right\}\:\mathrm{tan}\:{x} \\ $$$$\mathrm{28}+\mathrm{6}\sqrt{\mathrm{3}\:}\:=\:\mathrm{26}\:\mathrm{tan}\:{x} \\ $$$$\frac{\mathrm{14}+\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{13}}=\mathrm{tan}\:{x} \\ $$$${x}=\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{14}+\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{13}} \\ $$