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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.

1 Answers1

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$\DeclareMathOperator{\atanh}{arctanh}\DeclareMathOperator{\acsch}{arccosech}$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*}

Andrew D. Hwang
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