Given that, #a=cosx+isinx, i.e., a/1=(cosx+isinx)/1#.
Using componendo-didendo, we get,
#:. (a+1)/(a-1)=(cosx+isinx+1)/(cosx+isinx-1)#,
#=(cosx+1+isinx)/(cosx-1+isinx)xx(cosx-1-isinx)/(cosx-1-isinx)#,
#={(cosx+1+isinx)(cosx-1-isinx)}/{(cosx-1)^2-i^2sin^2x}#.
#:. (1+a)/(1-a)={(isinx+1+cosx)(isinx+1-cosx)}/{(cosx-1)^2+sin^2x}#,
#={(isinx+1)^2-cos^2x}/{(color(red)(cos^2x)-2cosx+color(red)1)+color(red)(sin^2x))#,
#={(i^2sin^2x+2isinx+1)-(cos^2x)}/(2-2cosx)#,
#=(color(blue)(-sin^2x)+2isinx+color(blue)1color(blue)(-cos^2x))/(2-2cosx)#,
#=(isinx)/(1-cosx)#,
#=(2isin(x/2)cos(x/2))/(2sin^2(x/2))#.
# rArr (1+a)/(1-a)=icot(x/2)#.
Feel the Joy of Maths.!