Question

If \(A, B, C, D\) be the angles of a cyclic quadrilateral, taken in order, prove that :

\[ \cos(180^\circ-A)+\cos(180^\circ+B)+\cos(180^\circ+C)-\sin(90^\circ+D)=0 \]

Solution

Using identities,

\[ \cos(180^\circ-A)=-\cos A \]

\[ \cos(180^\circ+B)=-\cos B \]

\[ \cos(180^\circ+C)=-\cos C \]

\[ \sin(90^\circ+D)=\cos D \]

Therefore,

\[ \begin{aligned} &\cos(180^\circ-A)+\cos(180^\circ+B) \\ &+\cos(180^\circ+C)-\sin(90^\circ+D) \\[8pt] =& -\cos A-\cos B-\cos C-\cos D \end{aligned} \]

In a cyclic quadrilateral,

\[ A+C=180^\circ, \qquad B+D=180^\circ \]

Hence,

\[ \cos C=-\cos A, \qquad \cos D=-\cos B \]

Substituting,

\[ \begin{aligned} &-\cos A-\cos B+\cos A+\cos B \\[8pt] =&0 \end{aligned} \]

Hence Proved.