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Question Number 206063 by MATHEMATICSAM last updated on 06/Apr/24

$$\mathrm{If}(x+\sqrt{1+x^2})(y+\sqrt{1+y^2})=1\mathrm{then}$$$$\mathrm{find}{(x+y)}^2.$$

Answered by mr W last updated on 06/Apr/24

$$x+\sqrt{1+x^2}=-y+\sqrt{1+{(-y)}^2}$$$$f\left(x\right)=x+\sqrt{1+x^2}isstrictlyincreasing.$$$$iff\left(a\right)=f\left(b\right),thena=b.$$$$\Rightarrow x=-y$$$$\Rightarrow x+y=0\Rightarrow {(x+y)}^2=0$$

Answered by Berbere last updated on 06/Apr/24

$$y=sh\left(t\right)=x=sh\left(w\right)$$$$(sh\left(w\right)+ch\left(w\right))(sh\left(t\right)+ch\left(t\right))=1$$$$e^w.e^t=1$$$$w+t=0$$$$x+y=sh\left(w\right)+sh(-w)=0$$

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