Question

If A, B, C, D are the angles of a cyclic quadrilateral, then $\cos A + \cos B + \cos{}C + \cos D =$

• A. $2(\cos A + \cos C)$
• B. $2(\cos A + \cos B)$
• C. $2(\cos A + \cos D)$
• D. 0

Answer 1

Answer: (D)

0

Answer 2

Given that ABCD is a cyclic quadrilateral.

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)

Adding, $\cos A + \cos B + \cos C + \cos D = 0.$