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In $\triangle A B C, \tan \frac{A}{2}+\tan \frac{B}{2}=$
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Verified Answer
The correct answer is:
$\frac{2 c \cot \frac{c}{2}}{a+b+c}$
$$
\begin{aligned}
& \text { }\left(\tan \frac{A}{2}+\tan \frac{B}{2}\right)=\frac{\Delta}{s(s-a)}+\frac{\Delta}{s(s-b)} \\
& =\frac{\Delta}{s}\left\{\frac{1}{s-a}+\frac{1}{s-b}\right\}=\frac{c \Delta}{s(s-a)(s-b)} \\
& =\frac{2 c \Delta}{(a+b+c)(s-a)(s-b)} \\
& =\frac{2 c}{(a+b+c)} \sqrt{\frac{s(s-c)}{(s-a)(s-b)}}=\frac{2 c \cot \frac{c}{2}}{a+b+c}
\end{aligned}
$$
\begin{aligned}
& \text { }\left(\tan \frac{A}{2}+\tan \frac{B}{2}\right)=\frac{\Delta}{s(s-a)}+\frac{\Delta}{s(s-b)} \\
& =\frac{\Delta}{s}\left\{\frac{1}{s-a}+\frac{1}{s-b}\right\}=\frac{c \Delta}{s(s-a)(s-b)} \\
& =\frac{2 c \Delta}{(a+b+c)(s-a)(s-b)} \\
& =\frac{2 c}{(a+b+c)} \sqrt{\frac{s(s-c)}{(s-a)(s-b)}}=\frac{2 c \cot \frac{c}{2}}{a+b+c}
\end{aligned}
$$
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