This is value of $\tan(z)$ in polar coordinates when I manually calculated it:
This is the value I got when I wrote the code for it in Maxima:
How can I reduce the first form to the second? I can't think of a way to convert the hyperbolic trigonometric function to normal ones. I tried doing this using the normal form of the formula $\tan(x+y)$, and I still don't seem to be gettting this equivalent form. Is there any other way to do it?
$$\tan(z)=\frac{\sin(x)\cosh(y)+i\cos(x)\sinh(y)}{\cos(x)\cosh(y)-i\sin(x)\sinh(y)}=$$
$\frac{[\sin(x)\cosh(y)+i\cos(x)\sinh(y)][\cos(x)\cosh(y)+i\sin(x)\sinh(y)]}{[\cos(x)\cosh(y)-i\sin(x)\sinh(y)][\cos(x)\cosh(y)+i\sin(x)\sinh(y)]}=$
$\frac{[\sin(x)\cosh(y)\cos(x)\cosh(y)-\cos(x)\sinh(y)\sin(x)\sinh(y)]+i[\cos(x)\sinh(y)\cos(x)\cosh(y)+\sin(x)\cosh(y)\sin(x)\sinh(y)]}{\cos(x)^2\cosh(y)^2+\sin(x)^2\sinh(y)^2}=$
$\frac{\sin(x)\cos(x)(\cosh(y)^2-\sinh(y)^2)+i\sinh(y)\cosh(y)(\cos(x)^2+\sin(x)^2)}{\cos(x)^2\cosh(y)^2+(1-\cos(x)^2)(\cosh(y)^2-1)}=\frac{\sin(x)\cos(x)+i\sinh(y)\cosh(y)}{\cos(x)^2+\cosh(y)^2-1}=$
$\frac{2\sin(x)\cos(x)+2i\sinh(y)\cosh(y)}{2\cos(x)^2+2\cosh(y)^2-2}=\frac{\sin(2x)+i\sinh(2y)}{(2\cos(x)^2-1)+(2\cosh(y)^2-1)}=\displaystyle{\frac{\sin(2x)+i\sinh(2y)}{\cos(2x)+\cosh(2y)}}$