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.
 A: 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 $$x=-y$$
A: Since
$$(x+\sqrt{x^2+1})(y+\sqrt{y^2+1})=1  / \cdot (x-\sqrt{x^2+1})$$
we get 
$$y+\sqrt{y^2+1}=\sqrt{x^2+1}-x$$ so
$$\sqrt{x^2+1}-\sqrt{y^2+1}=x+y\;/^2$$
and thus $$-\sqrt{y^2+1}\cdot \sqrt{x^2+1}+1= xy$$
so $$x^2y^2 +x^2+y^2+1=x^2y^2-2xy+1$$
and finaly $(x+y)^2=0$ so $x+y=0$
A: We have that $\mbox{arcsinh } x = \ln(x+\sqrt{x^2+1})$.  If we take logs of both sides of the equation we have 
$$\ln(x+\sqrt{x^2+1}) + \ln(y+\sqrt{y^2+1}) = 0$$
$$\mbox{arcsinh }x = -\mbox{arcsinh } y.$$
Then since $\mbox{arcsinh }$ is an odd function, we have $x=-y$.
A: Take natural logarithm to the both sides, and it becomes
$\ln(x+\sqrt{x^2+1})+ln(y+\sqrt{y^2+1})=0$
Which can be written as
$\sinh^{-1}x+\sinh^{-1}y=0$
Since $\sinh^{-1}$ is odd and strictly increasing, there is 
$\sinh^{-1}x=-\sinh^{-1}y=\sinh^{-1}-y$
Take $\sinh$ on both sides and Voila!
P.S. I didnt learn maths in English, so please tell me if I used a wrong word
A: 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) &= \cos\alpha\cos\beta(1-\sin\alpha)(1-\sin\beta) \\
\implies\cos^2\alpha\cos^2\beta &= \cos\alpha\cos\beta(1-\sin\alpha)(1-\sin\beta) \\
\implies \cos\alpha\cos\beta &= (1-\sin\alpha)(1-\sin\beta).
\end{align}
Thus
\begin{align} (1+\sin\alpha)(1+\sin\beta) = \cos\alpha\cos\beta &= (1-\sin\alpha)(1-\sin\beta) \\
\implies 1+\sin\alpha+\sin\beta+\sin\alpha\sin\beta &= 1-\sin\alpha-\sin\beta+\sin\alpha\sin\beta \\
\implies 2(\sin\alpha+\sin\beta) &= 0.
\end{align}
This implies that $\alpha = -\beta$, and hence $x = \tan\alpha = \tan(-\beta) = -\tan\beta = -y$, as desired.
