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Question Number 182715 by islamo last updated on 13/Dec/22

Commented by BaliramKumar last updated on 13/Dec/22

i think tan^(−1) (−x) = − tan^(−1) (x)

$${i}\:{think}\:{tan}^{−\mathrm{1}} \left(−{x}\right)\:=\:−\:{tan}^{−\mathrm{1}} \left({x}\right) \\ $$

Answered by Frix last updated on 13/Dec/22

tan y =±x ⇒   y=nπ+tan^(−1)  (±x) =nπ±tan^(−1)  x

$$\mathrm{tan}\:{y}\:=\pm{x}\:\Rightarrow\: \\ $$$${y}={n}\pi+\mathrm{tan}^{−\mathrm{1}} \:\left(\pm{x}\right)\:={n}\pi\pm\mathrm{tan}^{−\mathrm{1}} \:{x} \\ $$

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