if $(a+\sqrt{a^2+1})$ and $(b+\sqrt{b^2+1})$ are converse then prove that a and b are opposites $(a+\sqrt{a^2+1})\,(b+\sqrt{b^2+1})=1$ is supposed to equal: $b = -a$ but how do i get that? I've been trying to solve for like 2 days now.
 A: If
$$
(a+\sqrt{a^2+1})(b+\sqrt{b^2+1})=1$$
we have
$$
a+\sqrt{a^2+1}=\sqrt{b^2+1}-b \ \ \ \ \ (1)
$$
$$
b+\sqrt{b^2+1}=\sqrt{a^2+1}-a \ \ \ \ \ (2)
$$
$(1)+(2)$ then $a=-b$.
A: First approach. You can do it quite directly. This is not an elegant path, but it's easy to grasp and it is reliable. Regard your equality as an equation, where $b$ is fixed and $a$ is an unknown. How would you go about solving it?
It seems that the natural thing to do is to get $\sqrt{a^2+1}$ on one side of the equation, and then square everything to get rid of the radicals. Like this:
$$
\begin{align}
a + \sqrt{a^2+1} &= \frac{1}{\sqrt{b^2+1} + b} \\
\sqrt{a^2 + 1} &= \frac{1}{\sqrt{b^2+1} + b} - a \\
a^2 + 1 &= \left(\frac{1}{\sqrt{b^2+1} + b} - a\right)^2 = a^2 - \frac{2a}{\sqrt{b^2+1} + b} + \frac{1}{\left(\sqrt{b^2+1} + b\right)^2} \\
\frac{2a}{\sqrt{b^2+1}+b} &= \frac{1}{\left(\sqrt{b^2+1} + b\right)^2} - 1 \\
2a &= \frac{1}{\sqrt{b^2+1}+b} - \sqrt{b^2+1} - b = \frac{1 - (b^2 + 1) - 2b\sqrt{b^2+1} - b^2}{\sqrt{b^2+1} + b} = 
-2b
\end{align}
$$
And therefore $a=-b$.
Second approach. Another way to go about the problem is to exercise your curiosity. You might try to look at the converse statement, just to see what comes out of it: is it true that whenever $a=-b$, the equality $(a + \sqrt{a^2+1})(b + \sqrt{b^2+1})=1$ holds? It this is so, then it would mean that $(a + \sqrt{a^2+1})(-a + \sqrt{a^2+1})=1$ is true for any $a$. Does it look like that's indeed the case? Yes it does, by the formula for the difference of two squares:
$$
(\sqrt{a^2+1}+a)(\sqrt{a^2+1}-a) = (\sqrt{a^2+1})^2 - a^2 = 1.
$$
So, this is always true. Can we use this equality somehow to prove the original statement? We can, but I won't do it, because it's already been done in other answers.
