Simplify the following expression to its simplest form:

$$
\frac{\cot \alpha+\cot \beta}{\cot \alpha-\cot \beta}+\frac{\sin (\alpha+\beta)}{\sin (\alpha-\beta)}
$$
It can be easily shown for either of the two terms of our sum that it is equal to the other term with the opposite sign, and thus the value of our sum is 0.

For example,

$$
\frac{\operatorname{cotg} \alpha+\operatorname{cotg} \beta}{\operatorname{cotg} \alpha-\operatorname{cotg} \beta}=\frac{\frac{\cos \alpha}{\sin \alpha}+\frac{\cos \beta}{\sin \beta}}{\frac{\cos \alpha}{\sin \alpha}-\frac{\cos \beta}{\sin \beta}}=\frac{\cos \alpha \sin \beta+\cos \beta \sin \alpha}{\cos \alpha \sin \beta-\cos \beta \sin \alpha}=\frac{\sin (\alpha+\beta)}{\sin (\beta-\alpha)}
$$

But

$$
\sin (\beta-\alpha)=\sin [-(\alpha-\beta)]=-\sin (\alpha-\beta)
$$

and thus the value of our sum is:

$$
-\frac{\sin (\alpha+\beta)}{\sin (\alpha-\beta)}+\frac{\sin (\alpha+\beta)}{\sin (\alpha-\beta)}=0
$$

József Szabó (Szolnok, Beloiannisz g. III. o. t.)