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Question
In a ∆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
Solution
A, B, C and D are the angles of a cyclic quadrilateral.
\[ \therefore A + C = 180^\circ and B + D = 180^\circ\]
\[ \Rightarrow A = 180 - C and B = 180 - D\]
\[\text{ Now, LHS }= \cos\left( 180^\circ - A \right) + \cos\left( 180^\circ + B \right) + \cos\left( 180^\circ + C \right) - \sin\left( 90^\circ + D \right)\]
\[ = - \cos A + \left[ - \cos B \right] + \left[ - \cos C \right] - \cos D\]
\[ = - \cos A - \cos B - \cos C - \cos D\]
\[ = - \cos\left( 180^\circ - C \right) - \cos\left( 180^\circ - D \right) - \cos C - \cos D\]
\[ = - \left[ - \cos C \right] - \left[ - \cos D \right] - \cos C - \cos D\]
\[ = \cos C + \cos D - \cos C - \cos D\]
\[ = 0\]
= RHS
Hence proved.
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