First slide

Trigonometric Identities

Question

lf A,B,C and D are the angles of a cyclic quadrilateral, then cos A + cos B + cos C + cos D is equal to

Easy

A

2 (cos A + cos C)
B

2 (cos A + cos B)
C

2 (cos A + cos D)
D

0

Solution

Given that, ABCD is a cyclic quadrilateral.
$\begin{array}{ll} So, & A + C = 180^{\circ} \Rightarrow A = 180^{\circ} - C \\ \Rightarrow & \cos A = \cos (180^{\circ} - C) = - \cos C \\ \Rightarrow & \cos A + \cos C = 0 - - - (i) \\ Similarly, & \cos B + \cos D = 0 - - - - (ii) \end{array}$
On adding Eqs. (i) and (ii), we get
cosA + cosB + cosC + cosD =0

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