In Delta ABC,if tan frac{A}{2} tan frac{C}{2} | vedclass

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(D) Given $	an \frac{A}{2} 	an \frac{C}{2} = \frac{1}{2}$.Using the half-angle formulas $	an \frac{A}{2} = \sqrt{\frac{(s-b)(s-c)}{s(s-a)}}$ and $\

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None of these

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(D) Given $	an \frac{A}{2} 	an \frac{C}{2} = \frac{1}{2}$.Using the half-angle formulas $	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)}}$.Substituting these into the equation:$\sqrt{\frac{(s-b)(s-c)}{s(s-a)}} 	imes \sqrt{\frac{(s-a)(s-b)}{s(s-c)}} = \frac{1}{2}$$\sqrt{\frac{(s-b)^2}{s^2}} = \frac{1}{2}$$\frac{s-b}{s} = \frac{1}{2}$$2s - 2b = s$$s = 2b$Since $2s = a + b + c$,we have $a + b + c = 4b$,which implies $a + c = 3b$.This condition does not represent $A.P.$,$G.P.$,or $H.P.$.Therefore,the correct option is $D$.

In $\Delta ABC$,if $\tan \frac{A}{2} \tan \frac{C}{2} = \frac{1}{2}$,then $a, b, c$ are in

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