If A,B,C and D be the angles of a cyclic quadrilateral, taken in order, prove that

 cos(180^o-A) + cos(180^o+B) + cos(180^o+C) - sin(90^o+D) = 0

  -cosA-cosB-cosC-cosD

=-cosA-cosB-cos(180-A)-cos(180-B)

=-cosA-cosB+cosA+cosB

=0

please tell me whether my solution is correct or not?

In a cyclic quadrilateral, A+C=180 and B+D=180

                                         C=180-A               D = 180-B

       cos(180-A)+cos(180-B)+cos(180+C)+sin(90+D)    

  [cos(180-x)=-cosx, cos(180+x)=-cosx, sin(90+x)=-cosx]

       = -cosA-cosB-cosC-cosD

       =-(cosA+cosB+cosC+cosD)

       =-(cosA+cosB+cos(180-A)+cos(180-B))

       = - (cosA+cosB-cosA-cosB)

       = 0