Prove that if $({x+\sqrt{x^2+1}})({y+\sqrt{y^2+1}})=1$ then $x+y=0$ Let
$$\left({x+\sqrt{x^2+1}}\right)\left({y+\sqrt{y^2+1}}\right)=1$$
Prove that $x+y=0$.


This is my solution:
Let
$$a=x+\sqrt{x^2+1}$$
and
$$b=y+\sqrt{y^2+1}$$
Then $x=\dfrac{a^2-1}{2a}$ and $y=\dfrac{b^2-1}{2b}$. Now $ab=1\implies b=\dfrac1a$. Then I replaced $x$ and $y$:
$$x+y=\dfrac{a^2-1}{2a}+\dfrac{b^2-1}{2b}=\dfrac{a^2-1}{2a}+\dfrac{\dfrac{1}{a^2}-1}{\dfrac{2}{a}}=0$$
This solution is absolutely different from solution in my book. Is my solution mathematically correct? Did I assumed something that may not be true?
 A: Hint: Let $x=\sinh a$ and $y=\sinh b$. Then, using the fact that $\cosh^2u-\sinh^2u=1$ and $\sinh u+\cosh u=e^u$, we arrive at $e^{a+b}=1\iff a+b=0$, assuming a and b are reals.  Since $\sinh u$, just like $\sin u$, is an odd function, the proof is complete.
A: Note
$$y+\sqrt{y^2+1}=\sqrt{x^2+1}-x\tag{1}$$
$$x+\sqrt{x^2+1}=\sqrt{y^2+1}-y\tag{2}$$
$(1)+(2)$
$$\Longrightarrow x+y=-(x+y)$$
$$\Longrightarrow x+y=0$$
A: Given the symmetry of the problem, wlog $x\ge y$. $x=y=0$ is a solution, from now on let us assume that $x>0$.
Let $f$ be the real-valued map $w\mapsto w+\sqrt{w^2+1}$. If $w>0$ then $f(w)>1$, which is enough to prove that $x>0>y$. Even better (without deriving it), $f$ is strictly increasing on the positive reals.
Define $z=-y>0$. Then $f(x)f(y)=1$ is equivalent to (just rationalize it) 
$$f(x)=f(z),$$
which implies that $x=z$. This is equivalent to $x+y=0$.
A: Add other solution
since
$$\ln{(x+\sqrt{x^2+1})}+\ln{(y+\sqrt{y^2+1})}=0$$
note $f(x)=\ln{(x+\sqrt{x^2+1})}$ ia odd function and  strictly increasing on real numbers,
if $$f(x)+f(y)=0\Longleftrightarrow x+y=0$$
A: $x+\sqrt{x^2+1}$ is a strictly increasing, continuous function so that a single $x$ corresponds to a single $y$, and $y=-x$ is the one.
