If in a triangle ABC,acos^2frac{C}{2} + ccos^ | 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 formula $\cos^2\frac{A}{2} = \frac{s(s-a)}{bc}$ and $\cos^2\f

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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 formula $\cos^2\frac{A}{2} = \frac{s(s-a)}{bc}$ and $\cos^2\frac{C}{2} = \frac{s(s-c)}{ab}$,where $s = \frac{a+b+c}{2}$ is the semi-perimeter: $a \left( \frac{s(s-c)}{ab} \right) + c \left( \frac{s(s-a)}{bc} \right) = \frac{3b}{2}$ $\frac{s(s-c)}{b} + \frac{s(s-a)}{b} = \frac{3b}{2}$ $\frac{s}{b} (s-c + s-a) = \frac{3b}{2}$ Since $s-c+s-a = 2s - (a+c) = (a+b+c) - (a+c) = b$: $\frac{s}{b} (b) = \frac{3b}{2} \Rightarrow s = \frac{3b}{2}$ Substituting $s = \frac{a+b+c}{2}$: $\frac{a+b+c}{2} = \frac{3b}{2}$ $\Rightarrow a+b+c = 3b$ $\Rightarrow a+c = 2b$ Thus,$a, b, c$ are in $A.P.$

If in a triangle $ABC$,$a\cos^2\frac{C}{2} + c\cos^2\frac{A}{2} = \frac{3b}{2}$,then its sides will be in

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