$(x+\sqrt{x^2+1})(y+\sqrt{y^2+1})=1$ is true if and only if $x+y=0$ Prove that $(x+\sqrt{x^2+1})(y+\sqrt{y^2+1})=1$ is true if and only if $x+y=0$  
I believe x and y could both be 0 as that satisfies the equations. Beyond that, I do not know how to prove this.
 A: A short way:
Let $x=\sinh u$ and $y=\sinh v$. This turns the equation to $e^ue^v=1$.
As the hyperbolic sine is odd and invertible, $u=-v\iff x=-y$.

Another short way:
After simplification,
$$\left(\sqrt{x^2+1}-x+\sqrt{y^2+1}-y\right)\left(\left(\sqrt{x^2+1}+x\right)\left(\sqrt{y^2+1}+y\right)-1\right)\\
=2x+2y=0.$$
As the factor $\left(\sqrt{x^2+1}-x+\sqrt{y^2+1}-y\right)$ is guaranteed to be positive, the proof works both ways.
A: You have $$(x+\sqrt{x^2+1})(y+\sqrt{y^2+1}) = 1$$
$$x+\sqrt{x^2+1} = \sqrt{y^2+1} - y$$
$$x+y = -\sqrt{x^2+1} + \sqrt{y^2+1}$$
$$(x+y)(\sqrt{x^2+1} + \sqrt{y^2+1}) = -x^2+y^2$$
$$(x+y)(\sqrt{x^2+1} + \sqrt{y^2+1} +x-y) = 0$$
Now, you get $x+y=0$ or $\sqrt{x^2+1} + \sqrt{y^2+1} +x-y = 0$. Can you go further?
A: Hint:
$$x+\sqrt{1+x^2}=\sqrt{1+y^2}-y$$
WLOG $x=\cot2A,y=\cot2B$
$$x+\sqrt{1+x^2}=\cot A$$
$$\sqrt{1+y^2}-y=\tan B$$
$$\cot A=\tan B=\cot(90^\circ-B)$$
$A=180^\circ+90^\circ-B\implies2A=?,\cot2A=?$
A: By continuous rearranging and squaring:
$$\text{Equation} \Rightarrow x+\sqrt{x^2+1}=\frac{1}{y+\sqrt{y^2+1}} \Rightarrow x+y=\sqrt{y^2+1}-\sqrt{x^2+1} \Rightarrow$$
$$1-xy=\sqrt{x^2y^2+x^2+y^2+1} \Rightarrow (x+y)^2=0.$$
