If $ (x+\sqrt{x^2+1})(y+\sqrt{y^2+1})=1$, show that $x+y=0$ For $\{x,y\}\subset  \Bbb R$, $(x+\sqrt{x^2+1})(y+\sqrt{y^2+1})=1.$
Prove that $x+y=0.$

Problem presented in a book, as being from Norway Math Olympiad 1985. No answer was presented. My developments are not leading to a productive direction. Sorry if this is a duplicate. Hints and answers are welcomed.
Multiplying by $(x-\sqrt{x^2 +1})(y-\sqrt{y^2 +1})$ we get 

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

and thus $$(x+\sqrt{x^2 +1})(y+\sqrt{y^2 +1})=(x-\sqrt{x^2 +1})(y-\sqrt{y^2 +1})$$

hence 

$$x(\sqrt{x^2 +1} +\sqrt{y^2 +1} )=-y(\sqrt{x^2 +1} +\sqrt{y^2 +1} )$$
therefore $$x=-y$$
Let $x = \tan\alpha$ and $y = \tan\beta$ for some $\alpha,\beta\in(-\pi/2,\pi/2)$. Then $\sqrt{x^2+1} = \sec\alpha$ and $\sqrt{y^2+1} = \sec\beta$, so the condition can be rewritten as
$$(\tan\alpha+\sec\alpha)(\tan\beta+\sec\beta) = 1. $$
Multiplying by $\cos\alpha\cos\beta$ yields
$$(\sin\alpha+1)(\sin\beta+1) = \cos\alpha\cos\beta. $$
Multiplying by $(1-\sin\alpha)(1-\sin\beta)$ yields
\begin{align} (1-\sin^2\alpha)(1-\sin^2\beta) &amp;= \cos\alpha\cos\beta(1-\sin\alpha)(1-\sin\beta) \\
\implies\cos^2\alpha\cos^2\beta &amp;= \cos\alpha\cos\beta(1-\sin\alpha)(1-\sin\beta) \\
\implies \cos\alpha\cos\beta &amp;= (1-\sin\alpha)(1-\sin\beta).
\end{align}
Thus
\begin{align} (1+\sin\alpha)(1+\sin\beta) = \cos\alpha\cos\beta &amp;= (1-\sin\alpha)(1-\sin\beta) \\
\implies 1+\sin\alpha+\sin\beta+\sin\alpha\sin\beta &amp;= 1-\sin\alpha-\sin\beta+\sin\alpha\sin\beta \\
\implies 2(\sin\alpha+\sin\beta) &amp;= 0.
\end{align}
This implies that $\alpha = -\beta$, and hence $x = \tan\alpha = \tan(-\beta) = -\tan\beta = -y$, as desired.