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In $\triangle \mathrm{ABC},\left(\tan \frac{\mathrm{A}}{2}+\tan \frac{\mathrm{B}}{2}\right) \tan \frac{\mathrm{C}}{2}=$
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The correct answer is:
$\frac{2 c}{a+b+c}$
$\begin{aligned} & \text {}\left(\tan \frac{A}{2}+\tan \frac{B}{2}\right) \tan \frac{C}{2} \\ & =\tan \frac{A}{2} \tan \frac{C}{2}+\tan \frac{B}{2} \tan \frac{C}{2} \\ & =\sqrt{\frac{(s-b)(s-c)}{s(s-a)}} \sqrt{\frac{(s-a)(s-b)}{s(s-c)}} \\ & \quad+\sqrt{\frac{(s-a)(s-c)}{s(s-b)}} \sqrt{\frac{(s-a)(s-b)}{s(s-c)}} \\ & =\frac{s-b}{s}+\frac{s-a}{s}=\frac{2 s-a-b}{s} \\ & =\frac{a+b+c-a-b}{\left(\frac{a+b+c}{2}\right)}=\frac{2 c}{a+b+c}\end{aligned}$
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