Given $\left(x + \sqrt{1+y^2}\right)\left(y + \sqrt{1+x^2}\right) = 1$, prove $\left(x + \sqrt{1+x^2}\right)\left(y + \sqrt{1+y^2}\right) = 1$. 
Let $x$ and $y$ be real numbers such that 
  $$\left(x + \sqrt{1+y^2}\right)\left(y + \sqrt{1+x^2}\right) = 1$$ 
  Prove that 
  $$\left(x + \sqrt{1+x^2}\right)\left(y + \sqrt{1+y^2}\right) = 1$$


Hint :
By the hypothesis of the problem, do some algebra and show that :
$$\left(x + \sqrt{1+y^2}\right)\left(y + \sqrt{1+x^2}\right) = 1 \Leftrightarrow \dots \Leftrightarrow y = -x$$
Now, substitute $y=-x$ on the expression $\left(x + \sqrt{1+x^2}\right)\left(y + \sqrt{1+y^2}\right)$ and see what happens.
