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