# Express $\operatorname{sech}^{-1}(x)$ in terms of logarithms

I'm trying to express the following $$\operatorname{sech}^{-1}(x)$$ in terms of logarithms, and would warmly appreciate feedback towards my approach. The solution should be :

$$\ln\left(\dfrac{1+\sqrt{(1-x^2)}}{x}\right)$$ when $$x > 0.$$ However, I cannot seem to get this. My working out:

$$y = \operatorname{sech}^{-1}(x)\implies \operatorname{sech}(y) = x$$

$$-\operatorname{sech}(y)\operatorname{tan}(y) = x;$$

Once I got to here, I decided to transform the $$\operatorname{sech}$$ and $$\tan$$ into their hyperbolic identities.

$$-\dfrac{\operatorname{sinh}(y)}{\operatorname{cosh}^2(y)}=\dfrac{e^y-e^{-y}}{2}(\dfrac{e^y+e^{-y}}{2})^2 = x$$

Though I've tried reworking this into a quadratic form $$\dfrac{b \space \pm \space \sqrt{b^2-4ac}}{2a}$$, however, I couldn't manage to get the right form. I would greatly appreciate some help on the next steps towards this.

• Please use the mathjax basic tutorial and enhance your question. Feb 27, 2021 at 20:24
• Why not go first to $sech(y)=1/(\cosh(y))$ and then use definition of $\cosh(y)$ in terms of exponentials. Feb 27, 2021 at 20:33
• I don't see how you got $\tan$ here. Perhaps you meant $\operatorname{tanh}.$ In that case $-\operatorname{sech}(y)\operatorname{tanh}(y)=-x\sqrt{1-x^2}\neq x.$ Feb 27, 2021 at 20:39
• Reading through the answers, I realised that I was very close to this before the try above whilst practising on scrap-paper, but I made a simple arithmetic mistake that led me elsewhere. I must be getting tired! Feb 27, 2021 at 20:49

Once you have $$\operatorname{sech}(y) = x$$ it implies $$\dfrac{e^y+e^{-y}}2=\dfrac1x$$ which simplifies to the quadratic $$xe^{2y}-2e^y+x=0$$ of $$e^y.$$ Now solve for $$y$$ as you discribed in the last sentence.

By definition $$\operatorname{sech}(x)$$ is equal to $$y = \frac{2}{e^x+e^{-x}}.$$ What we now want to do is solve for $$x$$. First, divide both sides by $$2$$ and then take the multiplicative inverse of both sides to get $$\frac{2}{y}=e^x+e^{-x}$$ Now substitute $$u=e^x.$$ We will get a new and easy equation. $$u+\frac{1}{u}=\frac{2}{y}.$$ You should be able to solve this equation for $$u$$ $$u=\pm\sqrt{y^{-2}-1}+\frac{1}{y}.$$ If you are done, don't forget to substitute back $$e^x=u$$.

Forming a quadratic as outlined in the above answers is a fine approach, and one I recommend above this one. Here's a different one, which relies on knowing beforehand that $$\cosh^{-1}x=\ln(x+\sqrt{x^2-1})$$ Let $$y=\operatorname{sech} ^{-1} x$$. Then \begin{align} \operatorname{sech}y=x&\iff \cosh y=\frac{1}{x}\\ &\iff y=\operatorname{sech}^{-1}x=\cosh^{-1}\frac{1}{x}=\ln\left(\frac{1}{x}+\sqrt{\frac{1}{x^2}-1}\right) \end{align} Simplifying yields the correct answer. I hope that was helpful. If you have any questions please don't hesitate to ask :)

Note

\begin{align} \text{sech}\left(\ln\frac{1+\sqrt{1-x^2}}x \right) =& \frac2{ \exp\left(\ln\frac{1+\sqrt{1-x^2}}x \right)+\exp\left(-\ln \frac{1+\sqrt{1-x^2}}x\right)}\\ = &\frac2{ \frac{1+\sqrt{1-x^2}}x + \frac x{1+\sqrt{1-x^2}}} =\frac2{ \frac{1+\sqrt{1-x^2}}x + \frac{1-\sqrt{1-x^2}}x } =x \end{align}

Thus

$$\text{sech}^{-1}x =\ln\frac{1+\sqrt{1-x^2}}x$$

• This only shows that $\ln\left(\dfrac{1+\sqrt{(1-x^2)}}{x}\right)$ is a right inverse to $\operatorname{sech}(x).$ We also need to prove the other identity or bijectivity of one of these functions. Feb 28, 2021 at 1:44