Question

if a quadrilateral ABCD is cyclic ,then show that cos A +cosB +cosC +cosD = 0

Answer 1

hey mate,

As A,B,C,D are the vertices of quad ABCD,

A+B=180° and B+D=180°

By the rules of "association angle", we can write,

cosc = cos (2×90°-A)= -cosA ------------(1)

we can apply the same process on cosB or cosD -----------------

then, we shall get,

cosB= -cosD ---------------(2),

so, in L.H.S

= cosA+cosB+cosC+cosD

= cosA-cosD-cosA+cosD ------------ from 1 and 2,

= 0

hence, L.H.S=R.H.S

I would you suggest you to take a look at the concept of association angle.

I HOPE THIS HELPS YOU!