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Solve that following equations :`"tantheta"+"tan"(theta+pi/3)+"tan"(theta+(2pi)/3)=3`

Answer» Here, we will use, `tan(A+B) = (tanA+tanB)/(1-tanAtanB)`
It is given that ,
`tantheta+tan(theta+pi/3)+tan(theta+(2pi)/3) = 3`
`=>tantheta+(tantheta+tan(pi/3))/(1-tanthetatan((pi)/3)) +(tantheta+tan((2pi)/3))/(1-tanthetatan((2pi)/3)) = 3`
`=>(tantheta+sqrt3)/(1-sqrt3tantheta) +(tantheta-sqrt3)/(1+sqrt3tantheta) = 3-tantheta`
`=>((tantheta+sqrt3)(1+sqrt3tantheta) +(tantheta-sqrt3)(1-sqrt3tantheta))/(1-3tan^2theta) = 3-tantheta`
`=>(tantheta+sqrt3tan^2theta+sqrt3+3tantheta+tantheta-sqrt3tan^2theta-sqrt3+3tantheta)/(1-3tan^2theta) = 3-tantheta`
`=>(8tantheta)/(1-3tan^2theta) = 3-tantheta`
`=>8tantheta = (3-tantheta)(1-3tan^2theta)`
`=>8tantheta = 3-9tan^2theta-tantheta+3tan^2theta`
`=>9tantheta-3tan^3theta = 3-9tan^2theta`
`=>3tantheta - tan^3theta =1-3tan^2theta`
`=>(3tantheta - tan^3theta)/(1-3tan^2theta) = 1`
`=>tan3theta = tan(pi/4)`
`=>3theta = npi+pi/4`
`=>theta = (npi)/3+pi/12`


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