1. Determine which of the numbers is greater

$$
\operatorname{arctg}(2+\sqrt{5})+\operatorname{arcctg}(2-\sqrt{5}) \quad \text { or } \quad \frac{5 \sqrt{7}}{4} .
$$
# Problem 1.

Answer: the second number is greater.

Solution. If $a>0$ and $\varphi=\operatorname{arctg} a$, then $0<\varphi<\frac{\pi}{2}$ and $\operatorname{arctg} \frac{1}{a}=\frac{\pi}{2}-\varphi$. Therefore, $\operatorname{arctg} a + \operatorname{arctg} \frac{1}{a} = \frac{\pi}{2}$. If $a<0$, then $-\frac{1}{a}>0$, and $\operatorname{arctg} a + \operatorname{arctg} \left(-\frac{1}{a}\right) = -\frac{\pi}{2}$. Therefore, $\operatorname{arctg} a + \operatorname{arctg} \frac{1}{a} = -\frac{\pi}{2}$ if $a<0$.

If $a=\sqrt{2}+\sqrt{3}$, then $-\frac{1}{a}=\sqrt{2}-\sqrt{3}$, and if $a=2+\sqrt{5}$, then $-\frac{1}{a}=2-\sqrt{5}$; therefore, the first number equals $\pi$. Since $\sqrt{7}>2.6$, the second number: $\frac{5 \sqrt{7}}{4}>3.25>\pi$.