Question

cos A +2cos B+cos C=2. Prove that the sides are in A P

Answer 1

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(1−cosB)

⟹2cos(2A+C)cos(2A−C)=2(2sin22B)

⟹sin(2B)cos(2A−C)=2sin22B

⟹cos(2A−C)=2sin2B

Multiplying both sides by 2cos2B

we get 2cos2Bcos(

Therefore, a,b,c are in A.P.

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Answer 2

Answer:

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