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If $A, B, C, D$ are angles of a cyclic quadrilateral, then $\cos A+\cos B+\cos C+\cos D$ is equal to
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Verified Answer
The correct answer is:
$0$
We have,
$\angle A+\angle C=180^{\circ}$
and
$\angle B+\angle C=180^{\circ}$
$\begin{aligned} & \text { Now, } \cos A+\cos C+\cos B+\cos D \\ & =2 \cos \frac{A+C}{2} \cos \frac{A-C}{2}+2 \cos \frac{B+D}{2} \cos \frac{B-D}{2} \\ & =2 \cos 90^{\circ} \cos \frac{A-C}{2}+2 \cos 90^{\circ} \cos \frac{B-D}{2}=0\end{aligned}$
$\angle A+\angle C=180^{\circ}$
and
$\angle B+\angle C=180^{\circ}$
$\begin{aligned} & \text { Now, } \cos A+\cos C+\cos B+\cos D \\ & =2 \cos \frac{A+C}{2} \cos \frac{A-C}{2}+2 \cos \frac{B+D}{2} \cos \frac{B-D}{2} \\ & =2 \cos 90^{\circ} \cos \frac{A-C}{2}+2 \cos 90^{\circ} \cos \frac{B-D}{2}=0\end{aligned}$
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