need helphave to simplify and i don't know how (1+cos a-i*sin a)/(1-cos a+i*sina)

We'll re-write the sum of
numerator:

1+cos a = 2 [cos
(a/2)]^2

We'll re-write the sum of
denominator:

1-cos a = 2 [sin
(a/2)]^2

We'll re-write the
ratio:

(1+cos a-i*sin a)/(1-cos a+i*sina) = {2 [cos
(a/2)]^2-i*sin a}/{2 [sin (a/2)]^2+i*sina}

We'll factorize
and we'll get:

2 cos (a/2)*[cos (a/2) - i*sin (a/2)]/2 sin
(a/2)*[sin (a/2) + i*cos(a/2)]

We'll simplify and we'll
get:

cot (a/2)*{[cos (a/2) - i*sin (a/2)]/[sin (a/2) +
i*cos(a/2)]}

We'll multiply by the conjugate of
denominator:

cot (a/2)*{[cos (a/2) - i*sin (a/2)]*[sin
(a/2) - i*cos(a/2)]/{[sin (a/2)]^2 + [cos (a/2)]^2}}

But
{[sin (a/2)]^2 + [cos (a/2)]^2} = 1(fundamental formula of
trigonometry)

cot (a/2)*{[cos (a/2) - i*sin (a/2)]*[sin
(a/2) - i*cos(a/2)]

cot (a/2)*{cos (a/2)*sin (a/2) -
i*[cos(a/2)]^2 - i*[sin(a/2)]^2-cos (a/2)*sin (a/2)}

We'll
eliminate like terms:

-i*cot (a/2)* {[sin (a/2)]^2 + [cos
(a/2)]^2} = -i*cot (a/2)

The simplified ratio
is -i*cot (a/2).