$\underline{\text { Markelov C.B. }}$.

Can one, by applying to the number 2 the functions $\sin , \cos , \tan, \cot$, arcsin, arccos, arctan, $\operatorname{arccot}$ any number of times and in any order, obtain the number 2010?
The first method. When $\sin t > 0$ and $s \neq 0$, the identities $\sin t = \frac{1}{\sqrt{1 + \operatorname{ctg}^2 t}}$ and $\operatorname{ctg} \operatorname{arctg} \frac{1}{s} = s$ hold. Substituting $t = \operatorname{arcctg} 2$ and $s = \frac{1}{\sqrt{1 + \operatorname{ctg}^2 t}}$ into these identities, we get: $\operatorname{ctg} \operatorname{arctg} \sin \operatorname{arcctg} 2 = \sqrt{2^2 + 1}$. Similarly, $\operatorname{ctg} \operatorname{arctg} \sin \operatorname{arcctg} \sqrt{2^2 + 1} = 2^2 + 2, \ldots$ and after $2010^2 - 2^2$ repetitions of this operation, we obtain the number $\sqrt{2^2 + 1 + \ldots + 1} = 2010$.

Thus, $2010 = \underbrace{\operatorname{ctg} \operatorname{arctg} \sin \operatorname{arcctg}}_{2010^2 - 2^2} 2$, or, more concisely, $2010 = \operatorname{ctg} \underbrace{\operatorname{arctg} \sin}_{2010^2 - 2^2} \operatorname{arcctg} 2$.

$$
2010^2 - 2^2
$$

## Answer

It is possible.