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?

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 $$f(x)+f(y)=0\Longleftrightarrow x+y=0$$
