Verified answer

w = cos(2x) + i sin(2x)

DON'T FORGET PARENTHESES:

(w−1) / (w+1)

= (cos(2x) + i sin(2x) − 1) / (cos(2x) + i sin(2x) + 1)

= (cos(2x) + i sin(2x) − 1)(cos(2x) − i sin(2x) + 1) / [(cos(2x) + i sin(2x) + 1)(cos(2x) − i sin(2x) + 1)]

= [cos²(2x) − (i sin(2x) − 1)²] / [(cos(2x)+1)² − (i sin(2x))²]

= [cos²(2x) − (−sin²(2x) − 2i sin(2x) + 1)] / [(cos²(2x) + 2 cos(2x) + 1) − (−sin²(2x))]

= (2i sin(2x)) / (2 + 2 cos(2x))

= i sin(2x) / (1 + cos(2x))

= i * 2 sinx cosx / (1 + 2 cos²x − 1)

= i * 2 sinx cosx / (2 cos²x)

= i sinx/cosx

= i tanx

As written, the equation is false.

Perhaps you mean

.. (w-1)/(w+1) = i*tan(X)

Multiply by (w+1)*, i.e. the conjugate of w+1. Use w =cos(2X) +i*sin(2X).

.. = (w-1)(w+1)*/(2(cos(2X)+1))

.. = i*sin(2X)/(cos(2X)+1)

.. = i*tan(X) ... by the half-angle identity

w = cos 2x + i sin 2x

If you write the number in exponential form, it's even easier.

w = e^(2xi)

(w-1) / (w+1)

. = [e^(2xi) - 1] / [e^(2xi) + 1]

. = [e^(xi) - e^(-xi)] / [e^(xi) + e^(-xi)]

. = i * [ (e^(xi) - e^(-xi) ) / (2i) ] / [ (e^(xi) + e^(-xi)) / 2 ]

. = i * sin(x) / cos(x)

. = i * tan(x).