If `A ,B ,C ,D` are angles of a cyclic quadrilateral, then prove that `cos A+cos B+cos C+cos D=0`

If A ,\ B ,\ C ,\ D are angles of a cyclic quadrilateral, prove that cos A+cos B+cos C+cos D=0.

If A,B,C,D are angles of a cyclic quadrilateral,prove that cos A+cos B+cos C+cos D=0

If ABCD be a cyclic quadrilateral,-prove that cos A + cos B + cos C + cos D = 0

If A,B,C,D are the angles of a cyclic quadrilateral then cos A + cos B+ cos C + cos D=

I: If A,B,C,D are angles of a cyclic quadrilateral then cos A + cos B + cos C + cos D=0 II : If A,B,C,D are the angles of a quadrilateral then cos""(A+B)/(2) + cos""(C+D)/(2) =0

MA.B.C.D are the angles of a cyclic quadrilateral then cos 'fr cos B+cos C-cos D=

If A, B, C, D be the angles of a cyclic quadrilateral, taken in order, prove that : cos A + cos B + cos C+ cos D = 0 .

If A,B,C and D are the angles of cyclic quadrilateral, prove that: i) cosA + cosB+cosC+cosD = 0 ii) cos(180^(@)-A)+cos(180^(@)+B)+cos(180^(@)+C)-sin(90^(@)+D) = 0

If A,B,C and D are the angles of cyclic quadrilateral, prove that: i) cosA + cosB+cosC+cosD ii) cos(180^(@)-A)+cos(180^(@)+B)+cos(180^(@)+C)-sin(90^(@))