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Question Number 106286 by bemath last updated on 04/Aug/20

$$\mathrm{arc}\tan \left(\frac{\mathrm{x}+1}{\mathrm{x}-1}\right)+\mathrm{arc}\tan \left(\frac{\mathrm{x}-1}{\mathrm{x}}\right)=\mathrm{arc}\tan (-7)$$$$\mathrm{for}\mathrm{x}\mathrm{real}\mathrm{number}$$

Answered by bobhans last updated on 04/Aug/20

$$\to \mathrm{arc}\tan \left(\frac{\mathrm{x}+1}{\mathrm{x}-1}\right)+\mathrm{arc}\tan \left(\frac{\mathrm{x}-1}{\mathrm{x}}\right)=\mathrm{arc}\tan (-7)$$$$\tan (\mathrm{arc}\tan \left(\frac{\mathrm{x}+1}{\mathrm{x}-1}\right)+\mathrm{arc}\tan \left(\frac{\mathrm{x}-1}{\mathrm{x}}\right))=\tan \left(\mathrm{arc}\tan (-7)\right)$$$$\frac{\frac{\mathrm{x}+1}{\mathrm{x}-1}+\frac{\mathrm{x}-1}{\mathrm{x}}}{1-\frac{\mathrm{x}+1}{\mathrm{x}-1}.\frac{\mathrm{x}-1}{\mathrm{x}}}=-7\Rightarrow \frac{\mathrm{x}+1}{\mathrm{x}-1}+\frac{\mathrm{x}-1}{\mathrm{x}}=-7(1-\frac{{\mathrm{x}}^2-1}{{\mathrm{x}}^2-\mathrm{x}})$$$${\mathrm{x}}^2+\mathrm{x}+{\mathrm{x}}^2-2\mathrm{x}+1=-7({\mathrm{x}}^2-\mathrm{x}-{\mathrm{x}}^2+1)$$$${2\mathrm{x}}^2-\mathrm{x}+1=-7(-\mathrm{x}+1)$$$${2\mathrm{x}}^2-\mathrm{x}+1=7\mathrm{x}-7\Rightarrow {2\mathrm{x}}^2-8\mathrm{x}+8=0$$$$2{(\mathrm{x}-2)}^2=0\Rightarrow \mathrm{x}=2.★$$

Answered by Dwaipayan Shikari last updated on 04/Aug/20

$${tan}^{-1}\left(\frac{x+1}{x-1}\right)+{tan}^{-1}\left(\frac{x-1}{x}\right)$$$$={tan}^{-1}\left(\frac{\frac{x+1}{x-1}+\frac{x-1}{x}}{1-\frac{x+1}{x}}\right)={tan}^{-1}\left(\frac{\frac{x^2+x+x^2-2x+1}{(x-1)x}}{-\frac{1}{x}}\right)={tan}^{-1}(-7)$$$$\Rightarrow 2x^2-x+1=-7(1-x)$$$$\Rightarrow 2x^2-8x-8=0$$$$\Rightarrow x^2-4x+4=0\Rightarrow x=2$$

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