If A, B, C, D be

Given A, B, C and D are the angles of a cyclic quadrilateral.

∴ A + C = 180° and B + D = 180°

⇒ A = 180° – C and B = 180° - D

Now, LHS = cos (180^o – A) + cos (180^o + B) + cos (180^o + C) – sin (90^o + D)

= -cos A + [-cos B] + [-cos C] + [-cos D]

= -cos A – cos B – cos C – cos D

= -cos (180° - C) – cos (180° - D) – cos C – cos D

= -[-cos C] – [-cos D] – cos C – cos D

= cos C + cos D – cos C – cos D

= 0

= RHS

Hence proved.