If in a Delta ABC,a, b, c are in A.P.,then ta | vedclass

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(B) We know that $	an \frac{A}{2} 	an \frac{C}{2} = \sqrt{\frac{(s-b)(s-c)}{s(s-a)}} 	imes \sqrt{\frac{(s-a)(s-b)}{s(s-c)}} = \frac{s-b}{s}$.Since

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$\frac{1}{3}$

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(B) We know that $	an \frac{A}{2} 	an \frac{C}{2} = \sqrt{\frac{(s-b)(s-c)}{s(s-a)}} 	imes \sqrt{\frac{(s-a)(s-b)}{s(s-c)}} = \frac{s-b}{s}$.Since $a, b, c$ are in $A.P.$,we have $2b = a + c$.Adding $b$ to both sides,$3b = a + b + c = 2s$,so $b = \frac{2s}{3}$.Substituting this into the expression:$	an \frac{A}{2} 	an \frac{C}{2} = \frac{s - \frac{2s}{3}}{s} = \frac{\frac{s}{3}}{s} = \frac{1}{3}$.

If in a $\Delta ABC$,$a, b, c$ are in $A.P.$,then $\tan \frac{A}{2} \tan \frac{C}{2} = $

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