<s>[INST] Inverse hyperbolic functions are equal to each other but how I want to prove the following statement:
$${arctanh}(\frac{x}{\sqrt{1+x^2}})= {arccosech}(\frac{1}{x})$$
I consulted Schaum's Outlines of Mathematical Handbook of Formulas and Tables. It says:
$${arctanh}(x)=\frac{1}{2}ln(\sqrt{\frac{1+x}{1-x}})$$ 
$${arccosech}(x)=ln(\frac{1}{x}+\sqrt{\frac{1}{x^2}+1})$$
So I wrote my question in terms of those statements above:
$${arctanh}(\frac{x}{\sqrt{1+x^2}})= \frac{1}{2} ln(\frac{1+\frac{x}{\sqrt{1+x^2}}}{1-\frac{x}{\sqrt{1-x^2}}}) $$
$${arccosech}(\frac{1}{x})= ln(\frac{1}{\frac{1}{x}}+\sqrt{ \frac{1}{(\frac{1}{x})^2}+1})  $$
Now, the two equations those I must prove whether they are equal or not are:
$$ (\frac{1+\frac{x}{\sqrt{1+x^2}}}{1-\frac{x}{\sqrt{1-x^2}}})^\frac{1}{2} = (\frac{1}{\frac{1}{x}}+\sqrt{\frac{1}{(\frac{1}{x})^2}+1}) $$
Then I squared them both:
$$ \frac{\sqrt{1+x^2}+x}{\sqrt{1+x^2}}\frac{\sqrt{1-x^2}}{\sqrt{1-x^2}-x} = x^2 + 2x\sqrt{x^2+1}+x^2+1$$
But it seems to me it is going to anywhere. Any help or solving method is appreciated.
Regards.
 [/INST] $\DeclareMathOperator{\atanh}{arctanh}\DeclareMathOperator{\acsch}{arccosech}$Two key identities to note are
$$
1 \pm \frac{x}{\sqrt{1 + x^{2}}} = \frac{\sqrt{1 + x^{2}} \pm x}{\sqrt{1 + x^{2}}},
$$
so that
$$
\frac{1 + \dfrac{x}{\sqrt{1 + x^{2}}}}{1 - \dfrac{x}{\sqrt{1 + x^{2}}}}
= \frac{\sqrt{1 + x^{2}} + x}{\sqrt{1 + x^{2}} - x},
\tag{1}
$$
and the difference of squares factorization
$$
(\sqrt{1 + x^{2}} + x)(\sqrt{1 + x^{2}} - x) = (1 + x^{2}) - x^{2} = 1,
$$
which gives
$$
\sqrt{1 + x^{2}} + x = \frac{1}{\sqrt{1 + x^{2}} - x}.
\tag{2}
$$
Since $\atanh x = \ln \sqrt{\dfrac{1 + x}{1 - x}} = \dfrac{1}{2} \ln \dfrac{1 + x}{1 - x}$ (cf. your formula),
\begin{align*}
  \atanh \frac{x}{\sqrt{1 + x^{2}}}
  &= \frac{1}{2} \ln\frac{\sqrt{1 + x^{2}} + x}{\sqrt{1 + x^{2}} - x} && (1) \\
  &= \frac{1}{2} \ln (\sqrt{1 + x^{2}} + x)^{2} && (2) \\
  &= \ln (\sqrt{1 + x^{2}} + x) && \\
  &= \acsch \frac{1}{x}.
\end{align*}
</s>