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In a triangle $\mathrm{ABC}, \tan \frac{A}{2} \tan \frac{B}{2}+\tan \frac{B}{2} \tan \frac{C}{2}+\tan \frac{C}{2} \tan \frac{A}{2}=$
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$1$
$A+B+C=\pi$
$\begin{aligned} & \frac{A+B}{2}=\frac{\pi-C}{2} \\ & \frac{A}{2}+\frac{B}{2}=\frac{\pi}{2}-\frac{C}{2}\end{aligned}$
$\begin{aligned} \tan \left(\frac{A}{2}+\frac{B}{2}\right) & =\tan \left(\frac{\pi}{2}-\frac{C}{2}\right) \\ & =\cot \frac{C}{2} \ldots(\mathrm{i})\end{aligned}$
Also, $\tan \left(\frac{A}{2}+\frac{B}{2}\right)=\frac{\tan \frac{A}{2}+\tan \frac{B}{2}}{1-\tan \frac{A}{2} \tan \frac{B}{2}}$ ...(ii)
From (i) and (ii)
$\cot \frac{C}{2}=\frac{\tan \frac{A}{2}+\tan \frac{B}{2}}{1-\tan \frac{A}{2} \tan \frac{B}{2}}$
$\frac{1}{\tan \frac{c}{2}}=\frac{\tan \frac{A}{2}+\tan \frac{B}{2}}{1-\tan \frac{A}{2} \tan \frac{B}{2}}$
$1-\tan \frac{A}{2} \tan \frac{B}{2}=\tan \frac{B}{2} \tan \frac{C}{2}+\tan \frac{C}{2} \tan \frac{A}{2}$
$\therefore \tan \frac{A}{2} \tan \frac{B}{2}+\tan \frac{B}{2} \tan \frac{C}{2}+\tan \frac{C}{2} \tan \frac{A}{2}=1$
$\begin{aligned} & \frac{A+B}{2}=\frac{\pi-C}{2} \\ & \frac{A}{2}+\frac{B}{2}=\frac{\pi}{2}-\frac{C}{2}\end{aligned}$
$\begin{aligned} \tan \left(\frac{A}{2}+\frac{B}{2}\right) & =\tan \left(\frac{\pi}{2}-\frac{C}{2}\right) \\ & =\cot \frac{C}{2} \ldots(\mathrm{i})\end{aligned}$
Also, $\tan \left(\frac{A}{2}+\frac{B}{2}\right)=\frac{\tan \frac{A}{2}+\tan \frac{B}{2}}{1-\tan \frac{A}{2} \tan \frac{B}{2}}$ ...(ii)
From (i) and (ii)
$\cot \frac{C}{2}=\frac{\tan \frac{A}{2}+\tan \frac{B}{2}}{1-\tan \frac{A}{2} \tan \frac{B}{2}}$
$\frac{1}{\tan \frac{c}{2}}=\frac{\tan \frac{A}{2}+\tan \frac{B}{2}}{1-\tan \frac{A}{2} \tan \frac{B}{2}}$
$1-\tan \frac{A}{2} \tan \frac{B}{2}=\tan \frac{B}{2} \tan \frac{C}{2}+\tan \frac{C}{2} \tan \frac{A}{2}$
$\therefore \tan \frac{A}{2} \tan \frac{B}{2}+\tan \frac{B}{2} \tan \frac{C}{2}+\tan \frac{C}{2} \tan \frac{A}{2}=1$
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