Question

In cyclic quadrilateral $ABCD$, prove that
i . $\sin (A+B)+\sin (C+D)=0$
ii. $\cos (A+B)=\cos (C+D)$.

Answer 1

In cyclic quadrilateral
Angles $A+C=\pi ,B+D=\pi$

$\therefore \sin (A+B)=\sin (\pi -c+\pi -n)=\sin (2\pi -(C+D))=-\sin (C+D)$

$\Rightarrow \sin (A+B)+\sin (C+D)=0\cos (A+B)=\cos (\pi -C+\pi -D)=\cos (2\pi -(C+D))=\cos (C+D)$