Let's find the values of $x$ for which the functions

$$
y=\frac{1}{\tan x}+\frac{1}{\cot x} \quad \text{and} \quad y=\frac{1}{\tan x}-\frac{1}{\cot x}
$$

will have their minimum values.
1. 

$$
y=\frac{1}{\tan x}+\frac{1}{\cot x}=\frac{\cos x}{\sin x}+\frac{\sin x}{\cos x}=\frac{1}{\sin x \cos x}=\frac{2}{\sin 2 x}
$$

$y$ is minimal when $\sin 2 x$ is maximal, i.e., when $\sin 2 x=1$, which is when $2 x=90^{\circ}$, or $x=45^{\circ}$.

2. 

$$
y=\frac{1}{\tan x}-\frac{1}{\cot x}=\frac{\cos x}{\sin x}-\frac{\sin x}{\cos x}=\frac{\cos 2 x}{\sin x \cos x}=2 \cot 2 x
$$

Since the cotangent of an angle can take any value from $-\infty$ to $+\infty$, this function has neither a minimum nor a maximum.

(Pilczer Pál, Kaposvár.)

Solutions submitted: 39.