Question

1+cos alfa-isin alfa/1-cos alfa+isin alfa

Answer 1

[ 2cos²α/2 - 2isinα/2·cosα/2] / [ 2sin²α/2 +2isinα/2·cosα/2]=
  2cosα/2 ·[ cosα/2 - isinα/2] / 2sinα/2·[sinα/2  + icosα/2]=
   ctgα/2 · [  cos(-α/2)  + isin(-α/2) ] / [cosα/2 + isinα/2]
   ctgα/2 · [ cos( -α/2 -α/2)  + isin( -α/2 - α/2) ]
   ctgα/2 · [ cos( -α) + isin(-α)]
    ctgα/2 · [ cosα - isinα]
cosα = para  ; sinα=impara     ; cos(-α) = cosα  ; sin(-α) =- sinα