We use chain rule here. Let g(x)=cosx/(1+sinx) and f(x)=tan^(-1)x
then we can write h(x)=tan^(-1)(cosx/(1+sinx))
as h(x)=f(g(x))
and d/(dx)tan^(-1)(cosx/(1+sinx))
= d/(dx)f(g(x))
= (df)/(dg)xx(dg)/(dx)
Now as f(x)=tan^(-1)x, (df)/(dx)=1/(1+x^2)
and as g(x)=cosx/(1+sinx), using quotient rule
(dg)/(dx)=(-sinx(1+sinx)-cosx*cosx)/(1+sinx)^2
= (-sinx-sin^2x-cos^2x)/(1+sinx)^2=-1/(1+sinx)
Hence d/(dx)tan^(-1)(cosx/(1+sinx))
= 1/(1+(g(x))^2)xx(-1/(1+sinx))
= -1/(1+(cosx/(1+sinx))^2)xx1/(1+sinx)
= -(1+sinx)/((1+sinx)^2+cos^2x)
= -(1+sinx)/(1+sin^2x+2sinx+cos^2x)
= -1/2
Note: In case you are wondering why it turns out to be so simple answer, see the following.
cosx/(1+sinx)=(cos^2(x/2)-sin^2(x/2))/(cos(x/2)+sin(x/2))^2
= (cos(x/2)-sin(x/2))/(cos(x/2)+sin(x/2))
= (1-tan(x/2))/(1+tan(x/2))=(tan(pi/4)-tan(x/2))/(1+tan(pi/4)tan(x/2))=tan(pi/4-x/2)
hence tan^(-1)(cosx/(1+sinx))=pi/4-x/2 and its derivative is -1/2