Question

A,B,C and D are the angles of a cyclic quadrilateral. Prove that:

cosA+cosB+cosc+cosD=0
​

Answer 1

Answer:

As A,B,C,D are the angles of a cyclic quadrilateral.

so, angle A + angle C= 180°

= >angle A =180°-C

Cos A = -CosC

angle C= 180°- A

Cos C= - Cos A

silly angle B +angle D= 180°

angle B=180°- D

Cos B= -Cos D

angle D=180° -B

Cos D= -Cos B

Now,

L:H:S :

CosA + CosB + Cos C+ Cos D

= Cos A + Cos B + Cos ( 180°- A)+ Cos (180°- B)

= Cos A+Cos B-Cos A -Cos B

= 0(R:H:S)

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Answer 2

Answer:

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

A+C=180° and B+D=180°

By the rules of *associated angles*, we can write,

CosC=Cos(2x90-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&2

=0

Hence, L.H.S=R.H.S., |PROVED|

Step-by-step explanation:

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