Let's calculate (without a table) the value of the following expression:

$$
\operatorname{tg} \frac{\pi}{8}+\operatorname{ctg} \frac{\pi}{8}+\operatorname{tg} \frac{\pi}{16}-\operatorname{ctg} \frac{\pi}{16}+\operatorname{tg} \frac{\pi}{24}+\operatorname{ctg} \frac{\pi}{24}
$$
Based on known identities,

$$
\begin{aligned}
& \operatorname{tg} x+\operatorname{ctg} x=\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}=2 \sin 2 x \text { and } \\
& \operatorname{tg} x-\operatorname{ctg} x=-\frac{2 \cos 2 x}{\sin 2 x}=-2 \operatorname{ctg} 2 x
\end{aligned}
$$

According to these, the third and fourth terms of (1) together are equal to the second term multiplied by -2; thus, the first four terms together are equal to the difference of the first and second terms, which, according to (3), is equal to $-2 \operatorname{ctg} \pi / 4=-2$.

On the other hand, the last two terms of (1) together, according to (2), are equal to the reciprocal of half of $\sin \pi / 12$. Since $\pi / 12=\pi / 4-\pi / 6$, we have

$$
\sin \frac{\pi}{12}=\frac{1}{\sqrt{2}}\left(\frac{\sqrt{3}}{2}-\frac{1}{2}\right)=\frac{\sqrt{6}+\sqrt{2}}{4}=\frac{1}{\sqrt{6}+\sqrt{2}}
$$

Thus, the value of the given expression, rounded to four significant figures, is $-2+2(\sqrt{6}+\sqrt{2})=2(\sqrt{6}+\sqrt{2}-1) \approx 5.728$.

Janos Horanyi (Budapest, Apaczai Csere J. High School III. grade)

Note. Most of the solutions calculated all six terms of (1) separately, most often using the formulas

$$
\operatorname{tg} \frac{x}{2}=\sqrt{\frac{1-\cos x}{1+\cos x}}, \quad \operatorname{ctg} \frac{x}{2}=1 / \operatorname{tg} \frac{x}{2}, \quad \sin \frac{x}{2}=\sqrt{\frac{1-\cos x}{2}}
$$

In the given case, all the angles are acute angles, so the positive values of the square roots should be taken. It is more convenient to use the identity without the square root:

$$
\operatorname{tg} \frac{x}{2}=\frac{1-\cos x}{\sin x}=\frac{\sin x}{1+\cos x}
$$