<s>[INST] How to prove that $(x+ {\sqrt{1+x^2}} ) ( y+ {\sqrt {1+y^2}}) = 1$ if $(x+ {\sqrt{1+y^2}} ) ( y+ {\sqrt {1+x^2}}) = 1$ Let $x,y$ be real numbers such that :
$(x+ {\sqrt{1+y^2}} ) ( y+ {\sqrt {1+x^2}}) = 1$.
Prove that :
$(x+ {\sqrt{1+x^2}} ) ( y+ {\sqrt {1+y^2}}) = 1$.
I tried taking $x=y$. It simplifies everything a lot. But I'm not able to progress when both $x$ and $y$ are in the same equation.
 [/INST] Let $x =\sinh(a), y=\sinh(b)$. 
Then
$$(\sinh(a)+\cosh(b)) (\cosh(a) +\sinh(b))=1 \\
\sinh(a)\cosh(a)+\sinh(b)\cosh(b)+\cosh(a)\cosh(b)+\sinh(a)\sinh(b)=1 \\
\sinh(a+b)+\cosh(a+b)=1
$$
Now,
$$1=\cosh^2(a+b)-\sinh^2(a+b)=\left( \sinh(a+b)+\cosh(a+b)\right)\left( \cosh(a+b)-\sinh(a+b)\right) \\=\left( \cosh(a+b)-\sinh(a+b)\right)$$
Therefore
$$\sinh(a+b)+\cosh(a+b)=1\\
 \cosh(a+b)-\sinh(a+b)=1$$
 and hence 
$$\cosh(a+b)=1 \\
\sinh(a+b)=0$$
which proves that $a+b=0$. 
Therefore $y=-x$.
From here it is trivial.
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