Question

If $A$, $B$, $C$, $D$ be angles of a cyclic quadrilateral taken in order, then $cosA+cosB+cosC+cosD=$

• A. $0$
• B. $1$
• C. $-1$
• D. $1/2$

Answer 1

Answer: (A)

In a cyclic quadrilateral, the sum of opposite angles is $180^{\circ }$.
$\therefore A+C=180^{\circ }$ and $B+D=180^{\circ }$
$\Rightarrow C=180^{\circ }-A$ and $D=180^{\circ }-B\dots \left(i\right)$
Now, $cosA+cosB+cosC+cosD$
$=cosA+cosB+cos\left(180^{\circ }-A\right)+cos\left(180^{\circ }-B\right)$
[from $(i)$]
$=cosA+cosB-cosA-cosB=0$