If ABCD is a cyclic quadrilateral then `cos (180^@ +A) + cos(180^@-B )+cos(180^@-C)- sin (90^@-D)`

If ABCD is a cyclic quadrilateral then co (180^(@) +A) + cos (180^(@) - B) +cos(180^(@)-C) - sin (90^(@) - D)=

If ABCD is a cyclic quadrilateral then cos A + cos B + cos C + cos D is equal to

If A, B, C, D are the four angle taken in order of a cyclic quadrilateral, prove that (ii) cos (180^(@) - A) + cos (180^(@) + B) + cos (180^(@) + C) - sin (90^(@) + D) = 0

If ABCD is a cyclic quadrilateral, then the value of cos A - cos B + cos C - cos D =

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

If ABCD is a cyclic quadrilateral then the value of cosA+cosB+cos C+cos D is

If A ,\ B ,\ C ,\ D be the angles of a cyclic quadrilateral, taken in order, prove that: cos(180^0-A)+cos(180^0+B)+cos(180^0+C)-sin(90^0+D)=0

If ABCD is a cyclic quadrilateral,then the value of cos A-cos B+cos C-cos D is

If A, B, C and D be the angles of a cyclic quadrilateral, taken in order, prove that : cos (180^@- A) + cos (180^@+ B) + cos (180^@+ C)- sin (90^@ + D) = 0 .