If $a+\sqrt{a^2+1}= b+\sqrt{b^2+1}$, then $a=b$ or not?

Question

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.

Answer 1

Hint:

WLOG $a=\cot2A,b=\cot2B,0<A,B<\dfrac\pi2$

we have $\cot A=\cot B\implies\cot2A=?$

Answer 2

Hint: Square the equation $$a-b=\sqrt{b^2+1}-\sqrt{a^2+1}$$ two times. The result must be $$(a-b)^2=0$$

Answer 3

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 :

$$a + \sqrt{a^2+1} = b + \sqrt{b^2=1} \implies a=b$$

Answer 4

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, $$a+\sqrt{a^2+1}=b+\sqrt{b^2+1}\iff a=b.$$