In a ∆A, B, C, D be the angles of a cyclic quadrilateral,

$A, B, C$ 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.