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Question Number 191484 by MATHEMATICSAM last updated on 24/Apr/23

$$(x+\sqrt{1+x^2})(y+\sqrt{1+y^2})=1$$$$\mathrm{What}\mathrm{is}\mathrm{the}\mathrm{value}\mathrm{of}{(x+y)}^2?$$

Answered by witcher3 last updated on 24/Apr/23

$$\ln (\mathrm{x}+\sqrt{1+{\mathrm{x}}^2})+\ln (\mathrm{y}+\sqrt{1+{\mathrm{y}}^2)}=0$$$$\iff \mathrm{y}+\sqrt{{\mathrm{y}}^2+1}=\frac{1}{\mathrm{x}+\sqrt{1+{\mathrm{x}}^2}}=\sqrt{1+{\mathrm{x}}^2}-\mathrm{x}$$$$\Rightarrow \mathrm{let}\mathrm{f}\left(\mathrm{t}\right)=\mathrm{y}+\sqrt{1+{\mathrm{y}}^2}$$$${\mathrm{f}}^{\prime }\left(\mathrm{t}\right)=1+\frac{\mathrm{y}}{\sqrt{1+{\mathrm{y}}^2}}>0....\mathrm{E}$$$$\mathrm{we}\mathrm{have}\mathrm{f}\left(\mathrm{y}\right)=\mathrm{f}(-\mathrm{x})\Rightarrow \mathrm{withe}\mathrm{E}\mathrm{y}=-\mathrm{x}$$

Answered by mr W last updated on 24/Apr/23

$$x+\sqrt{1+x^2}=\sqrt{1+y^2}-y$$$$x+\sqrt{1+x^2}=-y+\sqrt{1+{(-y)}^2}$$$$\Rightarrow x=-y$$$$\Rightarrow x+y=0\Rightarrow {(x+y)}^2=0$$

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