<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] HINT:
Write $$x = \frac{1}{2}(s-\frac{1}{s})\\
y = \frac{1}{2}(t-\frac{1}{t})$$
with $s$, $t>0$. Then
$$( x+ \sqrt{1+y^2})(y + \sqrt{1+x^2}) -1 = (s t -1) \frac { s t (s+t)^2 + (s-t)^2}{ 4 s^2 t^2}$$
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