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If A, B, C, 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 |
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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(180°– A) + cos (180°+ B) + cos (180°+ C) – sin (90°+ 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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