In Delta ABC, if cos A + cos C = 4sin^2frac{B | vedclass

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(A) Given: $\cos A + \cos C = 4\sin^2\frac{B}{2}$Using the sum-to-product formula,$2\cos\frac{A+C}{2}\cos\frac{A-C}{2} = 4\sin^2\frac{B}{2}$Since $A+B

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$A.P.$

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(A) Given: $\cos A + \cos C = 4\sin^2\frac{B}{2}$Using the sum-to-product formula,$2\cos\frac{A+C}{2}\cos\frac{A-C}{2} = 4\sin^2\frac{B}{2}$Since $A+B+C = 180^{\circ},$ we have $\frac{A+C}{2} = 90^{\circ} - \frac{B}{2},$ so $\cos\frac{A+C}{2} = \sin\frac{B}{2}$Substituting this,$2\sin\frac{B}{2}\cos\frac{A-C}{2} = 4\sin^2\frac{B}{2}$$\cos\frac{A-C}{2} = 2\sin\frac{B}{2}$Using $\sin\frac{B}{2} = \cos\frac{A+C}{2},$ we get $\cos\frac{A-C}{2} = 2\cos\frac{A+C}{2}$$\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})$$3\sin\frac{A}{2}\sin\frac{C}{2} = \cos\frac{A}{2}\cos\frac{C}{2}$$	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)}},$ we get $\frac{s-b}{s} = \frac{1}{3}$$3s - 3b = s \implies 2s = 3b$$a+b+c = 3b \implies a+c = 2b$Thus,$a, b, c$ are in $A.P.$

In $\Delta ABC,$ if $\cos A + \cos C = 4\sin^2\frac{B}{2},$ then $a, b, c$ are in

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