In $ riangle A B C$, if $\frac{s-a}{11}=\frac{s-b}{12}=\frac{s-c}{13}$, then $ an ^2\left(\frac{A}{2} ight)+ an ^2\left(\frac{C}{2} ight)=$

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In $	riangle A B C$, if $\frac{s-a}{11}=\frac{s-b}{12}=\frac{s-c}{13}$, then $	an ^2\left(\frac{A}{2}ight)+	an ^2\left(\frac{C}{2}ight)=$

Aicharya · Accepted Answer

Given that, $\frac{s-a}{11}=\frac{s-b}{12}=\frac{s-c}{13}=k$ $ \begin{aligned} & s-a=11 k \ & s-b=12 k \ & s-c=13 k \end{aligned} $ On adding, $ \begin{aligned} 3 s-(a+b+c) & =36 k \ 3 s-2 s & =36 k \ s & =36 k \end{aligned} \quad[	ext { as } a+b+c=2 s] $ Now, $	an ^2 \frac{A}{2}+	an ^2 \frac{C}{2}$ $ \begin{aligned} & =\frac{(s-b)(s-c)}{s(s-a)}+\frac{(s-a)(s-b)}{s(s-c)} \ & =\frac{(12 k)(13 k)}{(36 k)(11 k)}+\frac{(11 k)(12 k)}{(36 k)(13 k)}=\frac{12}{36}\left[\frac{13}{11}+\frac{11}{13}ight] \ & =\frac{1}{3} \cdot\left[\frac{169+121}{11 	imes 13}ight]=\frac{290}{429} \end{aligned} $

In $\triangle A B C$, if $\frac{s-a}{11}=\frac{s-b}{12}=\frac{s-c}{13}$, then $\tan…

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