Answer :

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 (180o – A) + cos (180o + B) + cos (180o + C) – sin (90o + 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.


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