If $a+\sqrt{a^2+1}= b+\sqrt{b^2+1}$, then $a=b$ or not? It might be a silly question but if $$a+\sqrt{a^2+1}= b+\sqrt{b^2+1},$$ then can I conclude that $a=b$? I thought about squaring both sides but I think it is wrong! Because radicals will not be removed by doing that! Can you help me with proving that $a=b$ or not?
Actually I'm going to prove that $x+\sqrt{x^2+1}$ is a $1$-$1$ function.
 A: Hint: Square the equation
$$a-b=\sqrt{b^2+1}-\sqrt{a^2+1}$$ two times.
The result must be $$(a-b)^2=0$$
A: Let  $f:\mathbb R \to \mathbb R $ such that :
$$f(x)  = x + \sqrt{x^2+1}$$
Then, it is :
$$f'(x) = 1 + \frac{x}{\sqrt{x^2+1}}= \frac{\sqrt{x^2+1}+x}{\sqrt{x^2+1}} > 0, \; \forall x \in \mathbb R $$
Thus the function $f(x)$ is strictly increasing for $x \in \mathbb R$ and thus it is "$1-1$", which then means that :
$$a + \sqrt{a^2+1} = b +  \sqrt{b^2=1} \implies a=b$$
A: Alternatively, for $a,b\in \Bbb R$,
\begin{align}a+\sqrt{a^2+1}&=b+\sqrt{b^2+1}\\&\implies \frac{1}{a+\sqrt{a^2+1}}=\frac{1}{b+\sqrt{b^2+1}}\wedge a+\sqrt{a^2+1}=b+\sqrt{b^2+1}\\
&\implies \sqrt{a^2+1}-a=\sqrt{b^2+1}-b \wedge a+\sqrt{a^2+1}=b+\sqrt{b^2+1}\\
&\implies \sqrt{a^2+1}-a=\sqrt{b^2+1}-b \wedge a+\sqrt{a^2+1}=b+\sqrt{b^2+1}\\&\implies (a+\sqrt{a^2+1})-(\sqrt{a^2+1}-a)=(b+\sqrt{b^2+1})-(\sqrt{b^2+1}-b)
\\&\implies 2a=2b\implies a=b. \end{align}
It is also easy to see that $a=b\implies a+\sqrt{a^2+1}=b+\sqrt{b^2+1}$.  That is,
$$a+\sqrt{a^2+1}=b+\sqrt{b^2+1}\iff a=b.$$
A: Hint:
WLOG $a=\cot2A,b=\cot2B,0<A,B<\dfrac\pi2$
we have $\cot  A=\cot B\implies\cot2A=?$
