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


Write $x=\tan a$ and $y = \tan b$ for some (angles) $a,b$. They exsist since $\tan $ is surjective function. Pluging in starting equation we get:
$$ {\sin a+1\over \cos a}\cdot {\sin b+1\over \cos b}=1$$
and after rearranging we get $$\sin(a)+ \sin(b) = \cos (a+b) - \cos 0$$
which is equivalent to $$2\sin{a+b\over 2}\cos{a-b\over 2} = -2\sin {a+b\over 2}\sin{a+b\over 2}$$ 
Case 1. $\sin {a+b\over 2}=0$ then $a+b = 2\pi k$ so $$x=\tan a = \tan (2\pi k-b) = -\tan b = -y$$
Case 2. $\sin {a+b\over 2}\ne 0$, then $$\cos {a-b\over 2}+\sin{a+b\over 2}=0$$ 
Factorising this we get:
$$2\cos ({\pi\over 4}-{b\over 2})\cdot \cos({a\over 2}-{\pi\over 4})=0$$
Now we have to choises again.
1.st: $${\pi\over 4}-{b\over 2} = {\pi\over 2}+\pi k\implies b =-{\pi\over 2}+2\pi k $$
so $y$ does not exist.
2.nd case... we get $x$ does not exsist. 
So $x=-y$ and thus second expresins is also $1$.
