# 1. CONDITION

Calculate $\operatorname{arctg} 1+\operatorname{arctg} 2+\operatorname{arctg} 3$.
Solution. Since $\operatorname{arctg} 1=\pi / 4$, it is sufficient to compute $\operatorname{arctg} 2 +$ $\operatorname{arctg} 3$. From the formula for the tangent of the sum of angles, it follows that if $\alpha=\operatorname{arctg} 2, \beta=$ $\operatorname{arctg} 3$, then $\operatorname{tg}(\alpha+\beta)=(2+3) /(1-2 \cdot 3)=-1$. But, $\alpha, \beta \in(0 ; \pi / 2)$, so $0<\alpha+\beta$ $<\pi$, hence, $\alpha+\beta=3 \pi / 4$.

Remark. The problem can also be solved geometrically.