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

f(x)=arctan((√((x−1)/(2x+1)))) f^(−1) (x)=?

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

Commented by cortano1 last updated on 21/Oct/22

tan f(x)=(√((x−1)/(2x+1))) tan^2 f(x) = ((x−1)/(2x+1)) 2x tan^2 f(x)−x=−tan^2 f(x)−1 x=((−tan^2 f(x)−1)/(2tan^2 f(x)−1)) f^(−1) (y)=((−tan^2 y−1)/(2tan^2 y−1)) ∴ f^(−1) (x)=−((tan^2 x+1)/(2tan^2 x−1))

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

Commented by mathlove last updated on 21/Oct/22

thanks sir

$${thanks}\:{sir} \\ $$

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