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 $B=180-D$
Now, LHS $=\cos \left(180^{\circ}-A\right)+\cos \left(180^{\circ}+B\right)+\cos \left(180^{\circ}+C\right)-\sin \left(90^{\circ}+D\right)$
$=-\cos A+[-\cos B]+[-\cos C]-\cos D$
$=-\cos A-\cos B-\cos C-\cos D$
$=-\cos \left(180^{\circ}-C\right)-\cos \left(180^{\circ}-D\right)-\cos C-\cos D$
$=-[-\cos C]-[-\cos D]-\cos C-\cos D$
$=\cos C+\cos D-\cos C-\cos D$
$=0$
= RHS
Hence proved.