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Question Number 178743 by mathlove last updated on 21/Oct/22

$$f\left(x\right)=arctan\left(\sqrt{\frac{x-1}{2x+1}}\right)$$$$f^{-1}\left(x\right)=?$$

Commented by cortano1 last updated on 21/Oct/22

$$\tan \mathrm{f}\left(\mathrm{x}\right)=\sqrt{\frac{\mathrm{x}-1}{2\mathrm{x}+1}}$$$$\tan {}^2\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{x}-1}{2\mathrm{x}+1}$$$$2\mathrm{x}\tan {}^2\mathrm{f}\left(\mathrm{x}\right)-\mathrm{x}=-\tan {}^2\mathrm{f}\left(\mathrm{x}\right)-1$$$$\mathrm{x}=\frac{-\tan {}^2\mathrm{f}\left(\mathrm{x}\right)-1}{2\tan {}^2\mathrm{f}\left(\mathrm{x}\right)-1}$$$${\mathrm{f}}^{-1}\left(\mathrm{y}\right)=\frac{-\tan {}^2\mathrm{y}-1}{2\tan {}^2\mathrm{y}-1}$$$$\therefore {\mathrm{f}}^{-1}\left(\mathrm{x}\right)=-\frac{\tan {}^2\mathrm{x}+1}{2\tan {}^2\mathrm{x}-1}$$

Commented by mathlove last updated on 21/Oct/22

$$thankssir$$

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