In triangle ABC with the usual notations,if l | vedclass

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(C) We know that in $	riangle ABC$,$	an \frac{A}{2} = \sqrt{\frac{(s-b)(s-c)}{s(s-a)}}$ and $	an \frac{B}{2} = \sqrt{\frac{(s-a)(s-c)}{s(s-b)}}$.Gi

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(C) We know that in $	riangle ABC$,$	an \frac{A}{2} = \sqrt{\frac{(s-b)(s-c)}{s(s-a)}}$ and $	an \frac{B}{2} = \sqrt{\frac{(s-a)(s-c)}{s(s-b)}}$.Given $\left(	an \frac{A}{2}ight)\left(	an \frac{B}{2}ight)=\frac{3}{4}$.Substituting the formulas,we get $\sqrt{\frac{(s-b)(s-c)}{s(s-a)}} 	imes \sqrt{\frac{(s-a)(s-c)}{s(s-b)}} = \frac{3}{4}$.Simplifying the expression,$\sqrt{\frac{(s-c)^2}{s^2}} = \frac{3}{4}$,which gives $\frac{s-c}{s} = \frac{3}{4}$.Substituting $s = \frac{a+b+c}{2}$,we have $\frac{\frac{a+b+c}{2} - c}{\frac{a+b+c}{2}} = \frac{3}{4}$.This simplifies to $\frac{a+b-c}{a+b+c} = \frac{3}{4}$.Cross-multiplying gives $4(a+b) - 4c = 3(a+b) + 3c$.Therefore,$a+b = 7c$.

In $\triangle ABC$ with the usual notations,if $\left(\tan \frac{A}{2}\right)\left(\tan \frac{B}{2}\right)=\frac{3}{4}$,then $a+b=\ldots$ (in $c$)

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