In triangle ABC,if frac{s-a}{11}=frac{s-b}{12 | vedclass

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(A) Given that,$\frac{s-a}{11}=\frac{s-b}{12}=\frac{s-c}{13}=k$. $s-a=11k$,$s-b=12k$,$s-c=13k$. Adding these,we get $3s-(a+b+c) = 36k$. Since $a+b+c=2

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$\frac{290}{429}$

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(A) Given that,$\frac{s-a}{11}=\frac{s-b}{12}=\frac{s-c}{13}=k$. $s-a=11k$,$s-b=12k$,$s-c=13k$. Adding these,we get $3s-(a+b+c) = 36k$. Since $a+b+c=2s$,we have $3s-2s=36k$,so $s=36k$. Using the formula $	an^2\left(\frac{A}{2}ight) = \frac{(s-b)(s-c)}{s(s-a)}$ and $	an^2\left(\frac{C}{2}ight) = \frac{(s-a)(s-b)}{s(s-c)}$,$	an^2\left(\frac{A}{2}ight)+	an^2\left(\frac{C}{2}ight) = \frac{(12k)(13k)}{(36k)(11k)} + \frac{(11k)(12k)}{(36k)(13k)}$. $= \frac{12 	imes 13}{36 	imes 11} + \frac{11 	imes 12}{36 	imes 13} = \frac{1}{3} \left( \frac{13}{11} + \frac{11}{13} ight)$. $= \frac{1}{3} \left( \frac{169+121}{143} ight) = \frac{1}{3} 	imes \frac{290}{143} = \frac{290}{429}$.

In $\triangle ABC$,if $\frac{s-a}{11}=\frac{s-b}{12}=\frac{s-c}{13}$,then $\tan^2\left(\frac{A}{2}\right)+\tan^2\left(\frac{C}{2}\right) = $

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