How to show $\tanh(y\,\mathrm{i}) = \mathrm{i}\tan(y)$

Question

I have $\dfrac{e^{y\,\mathrm{i}}-e^{-y\,\mathrm{i}}}{e^{y\,\mathrm{i}}+e^{-y\,\mathrm{i}}}$, and I know that this equals $$ \frac{\frac{1}{2}(e^{y\,\mathrm{i}}-e^{-y\,\mathrm{i}})}{\frac{1}{2}(e^{y\,\mathrm{i}}+e^{-y\,\mathrm{i}})}=\frac{\sinh{y\,\mathrm{i}}}{\cosh{y\,\mathrm{i}}}=\tanh{y\,\mathrm{i}} $$ However, this is also supposed to equal $\mathrm{i}\tan y$. I'm not sure how to get to this result.

I tried doing $$ \frac{e^{y\,\mathrm{i}}-e^{-y\,\mathrm{i}}}{e^{y\,\mathrm{i}}+e^{-y\,\mathrm{i}}} = \frac{\frac{e^{2y\,\mathrm{i}}-1}{e^{y\,\mathrm{i}}}}{\frac{e^{2y\,\mathrm{i}}+1}{e^{y\,\mathrm{i}}}} = \frac{e^{2y\,\mathrm{i}}-1}{e^{2y\,\mathrm{i}}+1} $$ But this doesn't seem to have gotten me anywhere. Where should I go from here?

Answer 1

\begin{align}\sinh(yi)&=\frac{e^{yi}-e^{-yi}}2\\&=\frac12\left(\left(1+yi+\frac{(yi)^2}{2!}+\frac{(yi)^3}{3!}+\cdots\right)-\left(1-yi+\frac{(yi)^2}{2!}-\frac{(yi)^3}{3!}+\cdots\right)\right)\\&=\frac12\left(2yi-2\frac{y^3}{3!}i+2\frac{y^5}{5!}i-2\frac{y^7}{7!}i+\cdots\right)\\&=i\left(y-\frac{y^3}{3!}+\frac{y^5}{5!}-\frac{y^7}{7!}+\cdots\right)\\&=i\sin(y)\end{align}By a similar argument, $\cosh(yi)=\cos(y)$. Therefore, $\tanh(yi)=i\tan(y).$

Answer 2

Consider $$ \sinh(yi)= \frac{e^{yi}-e^{-yi}}{2}=i\frac{e^{yi}-e^{-yi}}{2i}=i\sin(y) $$ and $$ \cosh(yi)=\frac{e^{yi}+e^{-yi}}{2}=\cos(y) $$