The question

Express tan(1 - i) in the form a + ib, for a and b real

My attempt

I used the fact that:

sin(z) = sin(x)cosh(y) + icos(x)sinh(y)

cos(z) = cos(x)cosh(y) - isin(x)cosh(y)

I substituded these into tan(z) = \frac{sin(z)}{cos(z)}, and after quite a bit of algebra, got the following:

$\displaystyle \frac{sin(1)cos(1) - isinh(1)cosh(1)}{cos^2(1)cosh^2(1)+sin^2(1)sinh^2( 1)}$

I tried simplifying further, but I got a mess involving coth(1), cot(1) and other assorted trig functions. The solution in my text is:

$\displaystyle \frac{tan1^21 - itanh(1)sec^2(1)}{1 + tanh^2(1)tan^2(1)}$

Could someone guide me on how they got to that solution? Thank you.