Question

If A,B,C,D be the angles of a cyclic quadrilateral, taken in order, prove that: $cos({180}^{\circ }-A)+cos({180}^{\circ }+B)-sin({90}^{\circ }+D)=0$

Answer 1

$\because A,B,C,D$ are the angles of a cyclic quadrilateral in order,

$A+C=\pi andB+D=\pi$

$\Rightarrow \pi -A=Cand\pi -D=B$

$\Rightarrow cos(\pi -A)=cosC....\left(i\right)$

$\Rightarrow$ and cos $(\pi -D)=cosB....\left(ii\right)$

Now, $cos({180}^{\circ }-A)+cos({180}^{\circ }+B)+({180}^{\circ }+C)-sin({90}^{\circ }+D)$

$=cosC+(-cosB)-cosC-cosD$

$[\because cos({180}^{\circ }+B)=-cosB,cos({180}^{\circ }+C)=-cosC$ and using (i)]

=-cos B -cos D

=-cos B -(-cos B)[using (ii)]

=-cos B +cos B

=0

Hence proved.