Question Number 182715 by islamo last updated on 13/Dec/22
Commented by BaliramKumar last updated on 13/Dec/22
$${i}\:{think}\:{tan}^{−\mathrm{1}} \left(−{x}\right)\:=\:−\:{tan}^{−\mathrm{1}} \left({x}\right) \\ $$
Answered by Frix last updated on 13/Dec/22
$$\mathrm{tan}\:{y}\:=\pm{x}\:\Rightarrow\: \\ $$$${y}={n}\pi+\mathrm{tan}^{−\mathrm{1}} \:\left(\pm{x}\right)\:={n}\pi\pm\mathrm{tan}^{−\mathrm{1}} \:{x} \\ $$