In a triangle ABC with usual notations,if a, | vedclass

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(D) Given that $a, b, c$ are in arithmetic progression,we have $2b = a + c$. Using the half-angle formula for a triangle,$	an \frac{A}{2} = \sqrt{\fr

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$1/3$

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(D) Given that $a, b, c$ are in arithmetic progression,we have $2b = a + c$. Using the half-angle formula for a triangle,$	an \frac{A}{2} = \sqrt{\frac{(s-b)(s-c)}{s(s-a)}}$ and $	an \frac{C}{2} = \sqrt{\frac{(s-a)(s-b)}{s(s-c)}}$. Multiplying these,we get $	an \frac{A}{2} \cdot 	an \frac{C}{2} = \sqrt{\frac{(s-b)(s-c)}{s(s-a)} \cdot \frac{(s-a)(s-b)}{s(s-c)}} = \sqrt{\frac{(s-b)^2}{s^2}} = \frac{s-b}{s}$. Since $s = \frac{a+b+c}{2}$ and $a+c = 2b$,we have $s = \frac{2b+b}{2} = \frac{3b}{2}$. Substituting this into the expression,$\frac{s-b}{s} = \frac{\frac{3b}{2} - b}{\frac{3b}{2}} = \frac{\frac{b}{2}}{\frac{3b}{2}} = \frac{1}{3}$.

In a triangle $ABC$ with usual notations,if $a, b, c$ are in arithmetic progression,then $\tan \frac{A}{2} \cdot \tan \frac{C}{2} =$

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