In triangle ABC,if a, b, c are in A.P.,then t | vedclass

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(B) Given that $a, b, c$ are in $A.P.$,we have $2b = a + c$. Using the formula for $\sin \frac{A}{2} = \sqrt{\frac{(s-b)(s-c)}{bc}}$,$\sin \frac{C}{2}

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$1/2$

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(B) Given that $a, b, c$ are in $A.P.$,we have $2b = a + c$. Using the formula for $\sin \frac{A}{2} = \sqrt{\frac{(s-b)(s-c)}{bc}}$,$\sin \frac{C}{2} = \sqrt{\frac{(s-a)(s-b)}{ab}}$,and $\sin \frac{B}{2} = \sqrt{\frac{(s-a)(s-c)}{ac}}$. Substituting these into the expression: $\frac{\sin \frac{A}{2} \sin \frac{C}{2}}{\sin \frac{B}{2}} = \frac{\sqrt{\frac{(s-b)(s-c)}{bc}} 	imes \sqrt{\frac{(s-a)(s-b)}{ab}}}{\sqrt{\frac{(s-a)(s-c)}{ac}}} = \sqrt{\frac{(s-b)^2 (s-a)(s-c)}{ab^2c} 	imes \frac{ac}{(s-a)(s-c)}} = \sqrt{\frac{(s-b)^2}{b^2}} = \frac{s-b}{b}$. Since $s = \frac{a+b+c}{2}$ and $a+c = 2b$,we have $s = \frac{2b+b}{2} = \frac{3b}{2}$. Therefore,$\frac{s-b}{b} = \frac{\frac{3b}{2} - b}{b} = \frac{b/2}{b} = \frac{1}{2}$.

In triangle $ABC$,if $a, b, c$ are in $A.P.$,then the value of $\frac{\sin \frac{A}{2} \sin \frac{C}{2}}{\sin \frac{B}{2}} = $

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