Problem - 3873
In $\triangle{ABC}$ show that $$\tan\frac{A}{2}\tan\frac{B}{2}+\tan\frac{B}{2}\tan\frac{C}{2}+\tan\frac{C}{2}\tan\frac{A}{2}=1$$
Setting $C=\pi - (A - B)$ and also applying the sum of tangent formula:
\begin{align*}
&\tan\frac{A}{2}\tan\frac{B}{2}+\tan\frac{B}{2}\tan\frac{C}{2}+\tan\frac{C}{2}\tan\frac{A}{2}\\
&=\tan\frac{A}{2}\tan\frac{B}{2}+\tan\frac{C}{2}\Big(\tan\frac{A}{2}+\tan\frac{B}{2}\Big)\\
&=\tan\frac{A}{2}\tan\frac{B}{2}+\tan\frac{C}{2}\tan\Big(\frac{A}{2}+\frac{B}{2}\Big)\Big(1-\tan\frac{A}{2}\tan\frac{B}{2}\Big)\\
&=\tan\frac{A}{2}\tan\frac{B}{2}+\tan\frac{C}{2}\tan\Big(\frac{\pi}{2}-\frac{C}{2}\Big)\Big(1-\tan\frac{A}{2}\tan\frac{B}{2}\Big)\\
&=\tan\frac{A}{2}\tan\frac{B}{2}+\tan\frac{C}{2}\cot\frac{C}{2}\Big(1-\tan\frac{A}{2}\tan\frac{B}{2}\Big)\\
&=\tan\frac{A}{2}\tan\frac{B}{2}+\Big(1-\tan\frac{A}{2}\tan\frac{B}{2}\Big)\\
&=1
\end{align*}