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

Commented by BaliramKumar last updated on 13/Dec/22

$$ithink{tan}^{-1}(-x)=-{tan}^{-1}\left(x\right)$$

Answered by Frix last updated on 13/Dec/22

$$\tan y=\pm x\Rightarrow$$$$y=n\pi +{\tan }^{-1}(\pm x)=n\pi \pm {\tan }^{-1}x$$

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