In a triangle ABC,if a cos^2 frac{C}{2} + c c | vedclass

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(A) Given: $a \cos^2 \frac{C}{2} + c \cos^2 \frac{A}{2} = \frac{3b}{2}$ Using the half-angle formulas $\cos^2 \frac{C}{2} = \frac{s(s-c)}{ab}$ and $\c

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an arithmetic progression

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(A) Given: $a \cos^2 \frac{C}{2} + c \cos^2 \frac{A}{2} = \frac{3b}{2}$ Using the half-angle formulas $\cos^2 \frac{C}{2} = \frac{s(s-c)}{ab}$ and $\cos^2 \frac{A}{2} = \frac{s(s-a)}{bc}$: $a \cdot \frac{s(s-c)}{ab} + c \cdot \frac{s(s-a)}{bc} = \frac{3b}{2}$ $\frac{s(s-c)}{b} + \frac{s(s-a)}{b} = \frac{3b}{2}$ $\frac{s}{b} (s - c + s - a) = \frac{3b}{2}$ Since $2s = a + b + c$,we have $2s - a - c = b$: $\frac{s}{b} (b) = \frac{3b}{2}$ $s = \frac{3b}{2} \Rightarrow 2s = 3b$ Substituting $2s = a + b + c$: $a + b + c = 3b$ $a + c = 2b$ Thus,$a, b, c$ are in arithmetic progression.

In a $\triangle ABC$,if $a \cos^2 \frac{C}{2} + c \cos^2 \frac{A}{2} = \frac{3b}{2}$,then the sides of the triangle are in

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