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Question Number 125186 by Mathgreat last updated on 08/Dec/20

Answered by bemath last updated on 08/Dec/20

(1)arc tan ((1/3))+arc tan ((1/9)) =   arc tan ((((1/3)+(1/9))/(1−(1/(27)))))= arc tan (((12)/(26)))= arctan ((6/(13)))  (2) arc tan ((6/(13)))+arc tan ((7/(19)))=   arc tan ((((6/(13))+(7/(19)))/(1−(6/(13)).(7/(19))))) = arc tan (((205)/(205))) = (π/4)

$$\left(\mathrm{1}\right)\mathrm{arc}\:\mathrm{tan}\:\left(\frac{\mathrm{1}}{\mathrm{3}}\right)+\mathrm{arc}\:\mathrm{tan}\:\left(\frac{\mathrm{1}}{\mathrm{9}}\right)\:= \\ $$$$\:\mathrm{arc}\:\mathrm{tan}\:\left(\frac{\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{9}}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{27}}}\right)=\:\mathrm{arc}\:\mathrm{tan}\:\left(\frac{\mathrm{12}}{\mathrm{26}}\right)=\:\mathrm{arctan}\:\left(\frac{\mathrm{6}}{\mathrm{13}}\right) \\ $$$$\left(\mathrm{2}\right)\:\mathrm{arc}\:\mathrm{tan}\:\left(\frac{\mathrm{6}}{\mathrm{13}}\right)+\mathrm{arc}\:\mathrm{tan}\:\left(\frac{\mathrm{7}}{\mathrm{19}}\right)= \\ $$$$\:\mathrm{arc}\:\mathrm{tan}\:\left(\frac{\frac{\mathrm{6}}{\mathrm{13}}+\frac{\mathrm{7}}{\mathrm{19}}}{\mathrm{1}−\frac{\mathrm{6}}{\mathrm{13}}.\frac{\mathrm{7}}{\mathrm{19}}}\right)\:=\:\mathrm{arc}\:\mathrm{tan}\:\left(\frac{\mathrm{205}}{\mathrm{205}}\right)\:=\:\frac{\pi}{\mathrm{4}} \\ $$

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