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In $\triangle A B C,(a+b+c)\left(\tan \frac{A}{2}+\tan \frac{B}{2}\right)$ is equal to
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The correct answer is:
$2 c \cot \frac{C}{2}$
We have, $(a+b+c)\left(\tan \frac{A}{2}+\tan \frac{B}{2}\right)$
$\begin{aligned} & =2 s\left(\tan \frac{A}{2}+\tan \frac{B}{2}\right) \\ & =2\left(s \tan \frac{A}{2}+s \tan \frac{B}{2}\right) \\ & =2\left(\frac{\Delta}{s-a}+\frac{\Delta}{s-b}\right) \\ & =2 \Delta\left(\frac{s-b+s-a}{(s-a)(s-b)}\right) \\ & =2 \Delta \frac{2 s-(a+b)}{(s-a)(s-b)} \\ & =2 \Delta\left(\frac{c}{(s-a)(s-b)}\right) \\ & =2 c \cot \frac{C}{2}\end{aligned}$
$\begin{aligned} & =2 s\left(\tan \frac{A}{2}+\tan \frac{B}{2}\right) \\ & =2\left(s \tan \frac{A}{2}+s \tan \frac{B}{2}\right) \\ & =2\left(\frac{\Delta}{s-a}+\frac{\Delta}{s-b}\right) \\ & =2 \Delta\left(\frac{s-b+s-a}{(s-a)(s-b)}\right) \\ & =2 \Delta \frac{2 s-(a+b)}{(s-a)(s-b)} \\ & =2 \Delta\left(\frac{c}{(s-a)(s-b)}\right) \\ & =2 c \cot \frac{C}{2}\end{aligned}$
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