Question

If A,B, C, D are the angles of a cyclic quadrilateral show that cos(180-A)+cos(180+B)+cos(180+C)-sin(90+D)=0

Answer 1

Cos(180-A) = cos C as A+C= 190 degrees 
and cos(180+C) = -cosC 
Hence cos (180 - A) + cos (180 + B) + cos (180 + C) - sin (90 - D) = cos(180+B) - sin(90-D) 
sin(90-D) = sin(90-(180-B)) = sin(-(90-B)) = -sin (90-B) = -cosB where B is assumed acute angled as 90-B is considered positive 
hence cos(180+B) - sin(90-D) = -cosB - (-cosB) = 0

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Answer 2

Please see the attachment