Correct Answer - `A + iB = (1)/(2(1 + 3 cos^(2) (theta)/(2))) -I (cot (theta//2))/(1 + 3 cos^(2) (theta//2))`
Now `(1)/((1-cos theta)+2i sin theta)=(1)/(2 sin^2 ""theta/2+4i sin""theta/2 cos""theta/2)`
`=1/(2sin""theta/2(sin^2""theta/2+4 cos ^2""theta/2))xx(sin""theta/2-2 i cos ""theta/2)/((sin""theta/2-2 icos ""theta/2))`
`=(sin""theta/2-2i cos""theta/2)/(2 sin""theta/2(1 +3 cos^2""theta/2))`
`rArr A + iB =(1)/(2(1+3 cos^2""theta/2))-i(cot""theta/2)/(1+3 cos^2 ""theta/2)`
