How to deduce that $\arctan\left(\frac{a}{i}\right)=-i\cdot \text{ arctanh}(a)$

Question

I was doing some calculations in wolframalpha and I found the following equality:

$$\arctan\left(\frac{a}{i}\right)=-i\cdot \text{ arctanh}(a)$$

This is the first time I've seen this equality. How is this deduced? Is this just a specific case for $a/i$ or is it true also that:

$$\arctan\left(\frac{a}{b}\right)=-b\cdot \text{ arctanh}(a)?$$

Answer 1

First of all, note that $\frac ai=-ai$. So, the question becomes: is $\arctan(ai)=i\operatorname{arctanh}(a)$? Yes, because $\tan\bigl(\arctan(ai\bigr)=ai$ and$$\tan\bigl(i\operatorname{arctanh}(a)\bigr)=i\tanh\bigl(\operatorname{arctanh}(a)\bigr)=ia.\tag1$$The first equality from $(1)$ comes from$$\tan(ix)=\frac{\sin(ix)}{\cos(ix)}=\frac{i\sinh(x)}{\cosh(x)}=i\tanh(x).$$