In triangle ABC cosA+2cosB+cosC=2.Prove that sides of triangle are in A.P
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cos A + cos B + 2cos C = 2
=> cos A + cos B = 2 (1 - cos C)
=> (b²+c²-a²)/2bc + (c²+a²-b²)/2ca = 2[1 - (a²+b²-c²)/2ab]
=> a(b²+c²-a²) + b(c²+a²-b²) = 2c[2ab - (a²+b²-c²)]
=> ab² + ac² -a³ + bc² + ba² - b³ = 2c[2ab - (a²+b²-c²)]
=> ab² + ba² + ac² + bc² - a³ - b³ = 2c[2ab - (a²+b²-c²)]
=> ab(a+b) + c²(a+b) - (a+b)(a²-ab+b²) = 2c[2ab - (a²+b²-c²)]
=> (a+b)(ab+c² - a²+ab-b²) = 2c[2ab - (a²+b²-c²)]
=> (a+b)[2ab - (a²+b²-c²)] = 2c[2ab - (a²+b²-c²)]
=> a+b = 2c
=> sides of the triangle are in A.P.
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