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If $\alpha+\beta+\gamma=2 \pi$, then
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Verified Answer
The correct answer is:
$\tan \frac{\alpha}{2}+\tan \frac{\beta}{2}+\tan \frac{\gamma}{2}=\tan \frac{\alpha}{2} \tan \frac{\beta}{2} \tan \frac{\gamma}{2}$
$\begin{aligned} & \text { We have } \alpha+\beta+\gamma=2 \pi \Rightarrow \frac{\alpha}{2}+\frac{\beta}{2}+\frac{\gamma}{2}=\pi \\ & \Rightarrow \tan \left(\frac{\alpha}{2}+\frac{\beta}{2}+\frac{\gamma}{2}\right)=\tan \pi=0\end{aligned}$
$\begin{aligned} & \Rightarrow \tan \frac{\alpha}{2}+\tan \frac{\beta}{2}+\tan \frac{\gamma}{2}-\tan \frac{\alpha}{2} \tan \frac{\beta}{2} \tan \frac{\gamma}{2}=0 \\ & \Rightarrow \tan \frac{\alpha}{2}+\tan \frac{\beta}{2}+\tan \frac{\gamma}{2}=\tan \frac{\alpha}{2} \tan \frac{\beta}{2} \tan \frac{\gamma}{2}\end{aligned}$
$\begin{aligned} & \Rightarrow \tan \frac{\alpha}{2}+\tan \frac{\beta}{2}+\tan \frac{\gamma}{2}-\tan \frac{\alpha}{2} \tan \frac{\beta}{2} \tan \frac{\gamma}{2}=0 \\ & \Rightarrow \tan \frac{\alpha}{2}+\tan \frac{\beta}{2}+\tan \frac{\gamma}{2}=\tan \frac{\alpha}{2} \tan \frac{\beta}{2} \tan \frac{\gamma}{2}\end{aligned}$
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