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

Answer 1

$$\begin{gathered} A,B,C\mathrm{and}D\mathrm{are}\mathrm{the}\mathrm{angles}\mathrm{of}\mathrm{a}\mathrm{cyclic}\mathrm{quadrilateral}. \\ \therefore A+C=180°\mathrm{and}B+D=180° \\ \Rightarrow A=180-C\mathrm{and}B=180-D \\ \mathrm{Now},\mathrm{LHS}=\cos \left(180°-A\right)+\cos \left(180°+B\right)+\cos \left(180°+C\right)-\sin \left(90°+D\right) \\ =-\cos A+\left[-\cos B\right]+\left[-\cos C\right]-\cos D \\ =-\cos A-\cos B-\cos C-\cos D \\ =-\cos \left(180°-C\right)-\cos \left(180°-D\right)-\cos C-\cos D \\ =-\left[-\cos C\right]-\left[-\cos D\right]-\cos C-\cos D \\ =\cos C+\cos D-\cos C-\cos D \\ =0 \\ =\mathrm{RHS} \\ \mathrm{Hence}\mathrm{proved}. \end{gathered}$$