hyperbolic trig problem involving complex numbers

I have the following problem which Im trying to do from a text book:

If x + jy = inversetan (e^(a+jb)) then show that tan 2x = -cosb/ sinh a


Ive got thus far : getting rid of the inv tan by doing the tan of both sides

using the tan (a + b) identity for an expression for tan (x+jy) then using the identity tan jx = jtanhx.
this gives you :

tanx + jtanhy = e^(a+jb)*(1- j tanx tanhy)

by using the trig form of a complex number for e^(a+jb) which becomes e^a*(cosb+jsinb) you can multiply out the LHS to be able to equate the real and imaginery parts of the equation.
This gives you an expression for tanx equating the real parts.

tanx = e^a*(cosb+sinb tanx tanhy)

using tan 2x = 2tanx/ ( 1-(tanx)^2)

you can get an expression (a messy expression) for tan 2x but i cant simplify it further to get -cosb/sinha

Can anyone help?
Ive done the question twice once very carefully so i dont think ive made a mistake in my working out.