1+cosx/sin²x/(1+(1+cosx)/sinx)²) ! !

1+cosx/sin^2x/(1+(1+cosx)/sinx)^2)=

=2cos^2(x/2)/4sin^2(x/2)cos^2(x/2)/(1+2(1+cosx/sinx)+(1+cosx/sinx)^2)=

=1/2sin^2(x/2)/(1+2((1+cosx)/sinx)+(1+2cosx+cos^2x)/sin^2x)=

=1/2sin^2(x/2)/(1+2((1+cosx)/sinx)+((1+cosx)/sin^2x)+((cosx+cos^2x)/sin^2x)))=

=1/2sin^2(x/2)/(1+2((2cos^2(x/2))/4sin^(x/2)cos^2(x/2))+  +(2cos^2(x/2)/4sin^2(x/2)cos^2(x/2)+(cosx(1+cosx)/sin^2x))=

=1/2sin^2(x/2)/(1+(2cos(x/2)/sin(x/2)+(1/sin^2(x/2))+(cosx/2sin^2(x/2))=

=1/2sin^2(x/2)/((2sin^2(x/2)+4cos(x/2)sin(x/2)+2+cosx)/2sin^2x)=

=1/(2sin^2(x/2)+2sin2x+2+cosx)=1/(1-cosx+2sinx+cosx)=1/(1+2sinx)

ответ: 1/(1+2sinx)