Question

Anyone plzz answer this qstn:-If ABCD is a cyclic quadrilateral, then show that: cosA + cosB + cosC + cosD = 0​

Answer 1

$heya \\ \\ \\ \\ in \: acyclic \: quadrilateral \: \: \: \: \: a \: + c = 180 \\ \\ and \\ \\ \\ b + d = 180 \\ \\ \\ \\ \\ = > \\ \\ \\ cos(a) + \cos(180 - a) + \cos(d) + \cos(180 - d) \\ \\ \\ \cos(a) - cos(a) + \cos(d) - \cos(d) \\ \\ \\ = 0$

Answer 2

$\huge{\mathfrak{Question:-}$

If ABCD is a cyclic quadrilateral, then show that: cosA + cosB + cosC + cosD = 0​

$\huge{\mathfrak{Solution:-}$

In a quadrilateral, sum of opposite angles is 180°.

$\bold{A+C = 180^{\circ}}\\ \\ \bold{B+D = 180^{\circ}}$

L.H.S = cosA + cosB + cosC + cosD

$\bold{= (cos A + cos C)+(cosB + cosD)}\\ \\ \\ \bold{=2cos(\frac{A+C}{2}).\;cos(\frac{A-C}{2})+2cos(\frac{B+D}{2}).\;cos(\frac{B-D}{2})}\\ \\ \\ \bold{=2cos(\frac{180^{\circ}}{2}).\;cos(\frac{A-C}{2})+2cos(\frac{180^{\circ}}{2}).\;cos(\frac{B-D}{2})}\\ \\ \\ \bold{=2cos\;90^{\circ}.cos(\frac{A-C}{2})+2cos\;90^{\circ}.\;cos(\frac{B-D}{2})}\\ \\ \\ \bold{= 0\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;[\therefore cos\;90^{\circ}= 0]} \\ \\ \\ \bold{=R.H.S}\\ \\ \\ \bold{(HENCE\;PROVED)}$