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Q.
If $A$, $B$, $C$, $D$ be angles of a cyclic quadrilateral taken in order, then $cosA + cosB + cosC + cosD =$
Trigonometric Functions
Solution:
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 $D= 180^{\circ}-B\quad\ldots\left(i\right)$
Now, $cosA + cosB + cosC + cosD$
$= cosA + cosB + cos\left(180^{\circ} - A\right) + cos\left(180^{\circ} - B\right)$
[from $(i)$]
$= cosA + cosB - cosA - cosB = 0$