If in a Delta ABC,cos A + 2cos B + cos C = 2, | vedclass

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(A) Given: $\cos A + 2\cos B + \cos C = 2$Using the sum-to-product formula: $\cos A + \cos C = 2\cos\left(\frac{A+C}{2}\right)\cos\left(\frac{A-C}{2}\

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(A) Given: $\cos A + 2\cos B + \cos C = 2$Using the sum-to-product formula: $\cos A + \cos C = 2\cos\left(\frac{A+C}{2}ight)\cos\left(\frac{A-C}{2}ight)$Since $A+B+C = \pi$,we have $\frac{A+C}{2} = \frac{\pi}{2} - \frac{B}{2}$,so $\cos\left(\frac{A+C}{2}ight) = \sin\left(\frac{B}{2}ight)$Also,$2\cos B = 2(1 - 2\sin^2\frac{B}{2}) = 2 - 4\sin^2\frac{B}{2}$Substituting these into the equation: $2\sin\left(\frac{B}{2}ight)\cos\left(\frac{A-C}{2}ight) + 2 - 4\sin^2\left(\frac{B}{2}ight) = 2$$2\sin\left(\frac{B}{2}ight)\cos\left(\frac{A-C}{2}ight) = 4\sin^2\left(\frac{B}{2}ight)$Dividing by $2\sin\left(\frac{B}{2}ight)$ (since $\sin\left(\frac{B}{2}ight) 
eq 0$): $\cos\left(\frac{A-C}{2}ight) = 2\sin\left(\frac{B}{2}ight)$Substitute $\sin\left(\frac{B}{2}ight) = \cos\left(\frac{A+C}{2}ight)$: $\cos\left(\frac{A-C}{2}ight) = 2\cos\left(\frac{A+C}{2}ight)$Using $\cos(x-y) = \cos x \cos y + \sin x \sin y$ and $\cos(x+y) = \cos x \cos y - \sin x \sin y$:$\cos\frac{A}{2}\cos\frac{C}{2} + \sin\frac{A}{2}\sin\frac{C}{2} = 2(\cos\frac{A}{2}\cos\frac{C}{2} - \sin\frac{A}{2}\sin\frac{C}{2})$$3\sin\frac{A}{2}\sin\frac{C}{2} = \cos\frac{A}{2}\cos\frac{C}{2}$ $\Rightarrow 	an\frac{A}{2}	an\frac{C}{2} = \frac{1}{3}$Using $	an\frac{A}{2} = \sqrt{\frac{(s-b)(s-c)}{s(s-a)}}$,this simplifies to $a+c = 2b$,hence $a, b, c$ are in $A.P.$

If in a $\Delta ABC$,$\cos A + 2\cos B + \cos C = 2$,then $a, b, c$ are in

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