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Question

If in ABC, cosA+2cosB+cosC=2, prove that the sides of the triangle are in A.P.

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Solution

Given that A,B,C are the angles of a triangle.

Therefore, A+B+C=180

Given that cosA+2cosB+cosC=2

cosA+cosC=2(1cosB)

2cos(A+C2)cos(AC2)=2(2sin2B2)

sin(B2)cos(AC2)=2sin2B2

cos(AC2)=2sinB2

Multiplying both sides by 2cosB2

we get 2cosB2cos(AC2)=2(2cosB2sinB2)

2cos180(A+C)2cos(AC2)=2sinB

2sin(A+C2)cos(AC2)=2sinB

sinA+sinC=2sinB

a+c=2b

Therefore, a,b,c are in A.P.

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