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Question Number 145190 by qaz last updated on 03/Jul/21

$$\mathrm{f}\left(\mathrm{x}\right)+\mathrm{f}(1-\frac{1}{\mathrm{x}})={\tan }^{-1}\mathrm{x},(\mathrm{x}\ne 0)$$$$\mathrm{Find}\mathrm{f}\left(\mathrm{x}\right)=?$$

Answered by EDWIN88 last updated on 03/Jul/21

$$\left(1\right)f\left(x\right)+f\left(\frac{x-1}{x}\right)={\tan }^{-1}\left(x\right)$$$$replacex\to \frac{x-1}{x}$$$$\left(2\right)f\left(\frac{x-1}{x}\right)+f\left(\frac{\frac{x-1}{x}-1}{\frac{x-1}{x}}\right)={\tan }^{-1}\left(\frac{x-1}{x}\right)$$$$\Rightarrow f\left(\frac{x-1}{x}\right)+f\left(\frac{1}{1-x}\right)={\tan }^{-1}\left(\frac{x-1}{x}\right)$$$$replacex\to \frac{1}{1-x}$$$$\left(3\right)f\left(\frac{1}{1-x}\right)+f\left(\frac{\frac{1}{1-x}-1}{\frac{1}{1-x}}\right)={\tan }^{-1}\left(\frac{1}{1-x}\right)$$$$\Rightarrow f\left(\frac{1}{1-x}\right)+f\left(x\right)={\tan }^{-1}\left(\frac{1}{1-x}\right)$$$$sumequation\left(1\right),\left(2\right),\left(3\right)$$$$\Rightarrow 2[f\left(x\right)+f\left(\frac{1}{1-x}\right)+f\left(\frac{x-1}{x}\right)]={\tan }^{-1}\left(x\right)+{\tan }^{-1}\left(\frac{x-1}{x}\right)+{\tan }^{-1}\left(\frac{1}{1-x}\right)$$$$\Rightarrow 2[f\left(x\right)+{\tan }^{-1}\left(\frac{x-1}{x}\right)]={\tan }^{-1}\left(x\right)+{\tan }^{-1}\left(\frac{x-1}{x}\right)+{\tan }^{-1}\left(\frac{1}{1-x}\right)$$$$2f\left(x\right)={\tan }^{-1}\left(x\right)+{\tan }^{-1}\left(\frac{1}{1-x}\right)-{\tan }^{-1}\left(\frac{x-1}{x}\right)$$$$f\left(x\right)=\frac{1}{2}[{\tan }^{-1}\left(x\right)+{\tan }^{-1}\left(\frac{1}{1-x}\right)-{\tan }^{-1}\left(\frac{x-1}{x}\right)]$$$$=\frac{1}{2}[{\tan }^{-1}\left(\frac{x+\frac{1}{1-x}}{1-\frac{x}{1-x}}\right)-{\tan }^{-1}\left(\frac{x-1}{x}\right)]$$$$=\frac{1}{2}[{\tan }^{-1}\left(\frac{x+1-x^2}{1-2x}\right)-{\tan }^{-1}\left(\frac{x-1}{x}\right)]$$$$=\frac{1}{2}{\tan }^{-1}\left(\frac{\frac{x+1-x^2}{1-2x}-\frac{x-1}{x}}{1+\frac{(x+1-x^2)(x-1)}{(1-2x)x}}\right)$$$$=\frac{1}{2}{\tan }^{-1}\left(\frac{x^2+x-x^3-(1-2x)(x-1)}{1-2x^2+(x-1)(x+1-x^2)}\right)$$

Commented by qaz last updated on 03/Jul/21

$$\mathrm{thank}\mathrm{you}$$

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