Q. If A,B,C,D are the angles of a cyclic quadrilateral, then $\cos{A + \cos{B + \cos{D\ }}}$is equal to
A) $2(\cos{A + \cos{C)}}$
B) $2(\cos{A + \cos{B)}}$
C) $2(\cos{A + \cos{D)}}$
D) 0
Explanation
(d)
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, $\cos{B + \cos{D = 0\ldots(ii)}}$
On adding Eqs. (i) and (ii), we get
$\cos{A + \cos{B + \cos{C + \cos{D = 0}}}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.