TITLE: Calculate $\tan(1+i)$
QUESTION [5 upvotes]: Calculate $\tan(1+i)$.
I use the expression $\tan x = -i\dfrac{e^{ix}-e^{-ix}}{e^{ix}+e^{-ix}}$. So it only remains to calculate $e^{i(1+i)}$ (and then $e^{-i(1+i)}$ follows by taking the reciprocal). 
So $e^{i(1+i)} = e^ie^{ii} = e^i/e$. 
For $e^i$ I use the formula $e^x = 1+x+\dfrac{x^2}{2!}+\ldots$, so $$e^i = 1+i-\dfrac{1}{2!}-\dfrac{i}{3!}+\dfrac{1}{4!} = \left(1-\dfrac{1}{2!}+\dfrac{1}{4!}-\ldots\right)+i\left(1-\dfrac{1}{3!}+\dfrac{1}{5!}-\ldots\right)$$
and there's probably no way to calculate that except say that it's equal to $\cos 1+i\sin 1$.
So $e^i/e = (\cos 1+i\sin 1)/e$, and $e/e^i = e/(\cos 1+i\sin 1)$.
This seems to yield a very ugly expression for the original $\tan(1+i)$. Is there any way to simplify?

REPLY [2 votes]: HINT:
Let $a+ib=\tan(1+i),\implies a-ib=\tan(1-i)$
So using $2\cos A\cos B=\cos(A-B)+\cos(A+B)$ and $\cos(ix)=\cosh x,\sin(ix)=i\sinh x$
$$2a=\tan(1+i)+\tan(1-i)=\frac{\sin2}{\cos(1+i)\cos(1-i)}=\frac{2\sin2}{\cos2i+\cos2}=\frac{2\sin2}{\cosh2+\cos2}$$
$$2ib=\tan(1+i)-\tan(1-i)=\frac{\sin2i}{\cos(1+i)\cos(1-i)}=\frac{2i\sinh2}{\cos2i+\cos2}=\frac{2i\sinh2}{\cosh2+\cos2}$$