22. Find the value of $\frac{\tan 40^{\circ} \tan 60^{\circ} \tan 80^{\circ}}{\tan 40^{\circ}+\tan 60^{\circ}+\tan 80^{\circ}}$.
22. Answer: 1
We show that more generally, if $\triangle \mathrm{ABC}$ is acute-angled, then
$$
\frac{\tan A \cdot \tan B \cdot \tan C}{\tan A+\tan B+\tan C}=1 .
$$

We have
$$
\begin{aligned}
& \tan A+\tan B+\tan C \\
= & \tan A+\tan B+\tan \left(180^{\circ}-(A+B)\right) \\
= & \tan A+\tan B-\tan (A+B) \\
= & \tan A+\tan B-\frac{\tan A+\tan B}{1-\tan A \tan B} \\
= & (\tan A+\tan B)\left(1-\frac{1}{1-\tan A \cdot \tan B}\right) \\
= & (\tan A+\tan B)\left(\frac{-\tan A \cdot \tan B}{1-\tan A \cdot \tan B}\right) \\
= & (\tan A \cdot \tan B)\left(-\frac{\tan A+\tan B}{1-\tan A \cdot \tan B}\right) \\
= & \tan A \cdot \tan B \cdot \tan \left(180^{\circ}-(A+B)\right) \\
= & \tan A \cdot \tan B \cdot \tan C,
\end{aligned}
$$
so the result follows.