What is the real part of the complex number (1-cosx-i*sinx)/(1+cosx+i*sinx)?

We'll re-write the numerator and denominator of the given
complex number, using the half angle
identities:

(1-cosx-i*sinx)/(1+cosx+i*sinx) = {2[sin
(x/2)]^2 - i*sin x}/{2[cos (x/2)]^2 + i*sin x}

We'll
re-write the term sin x, using the double angle
identity:

sin x = 2 sin(x/2)*cos
(x/2)

(1-cosx-i*sinx)/(1+cosx+i*sinx) = {2[sin (x/2)]^2 -
i*2 sin(x/2)*cos (x/2)}/{2[cos (x/2)]^2 + i*2 sin(x/2)*cos
(x/2)}

We'll factorize and we'll
get:

(1-cosx-i*sinx)/(1+cosx+i*sinx) = 2 sin(x/2){[sin
(x/2)] - i*cos (x/2)}/2cos (x/2){[cos (x/2)] + i*2
sin(x/2)}

We'll simplify and we'll
get:

(1-cosx-i*sinx)/(1+cosx+i*sinx) = 2 sin(x/2){[sin
(x/2)] - i*cos (x/2)}/2cos (x/2){[cos (x/2)] + i*2
sin(x/2)}

(1-cosx-i*sinx)/(1+cosx+i*sinx) = 2 sin(x/2){[cos
(x/2)] + i*sin (x/2)}/2cos (x/2)*i*{[cos (x/2)] + i*2
sin(x/2)}

(1-cosx-i*sinx)/(1+cosx+i*sinx) =
tan(x/2)/i

Since it is not allowed to keep a complex number
at denominator, we'll multiply by i both numerator and
denominator.

(1-cosx-i*sinx)/(1+cosx+i*sinx) =
i*tan(x/2)/i^2

But i^2 =
-1

(1-cosx-i*sinx)/(1+cosx+i*sinx) =
-i*tan(x/2)

The rectangular form of a complex number
is:

z = x + i*y, where Re(z) = x and Im(z) =
y

Comparing both forms, we'll get Re(z) = 0 and Im(z) = -
tan(x/2).

Therefore, the real part of the
given complex number is Re(z) = 0.