3.171. Given: $\operatorname{ctg} \alpha=\frac{3}{4}, \operatorname{ctg} \beta=\frac{1}{7}, 0<\alpha<\frac{\pi}{2}, 0<\beta<\frac{\pi}{2}$. Find $\alpha+\beta$.
## Solution.

$$
\begin{aligned}
& \operatorname{ctg} \alpha+\operatorname{ctg} \beta=\frac{3}{4}+\frac{1}{7}, \quad \frac{\cos \alpha}{\sin \alpha}+\frac{\cos \beta}{\sin \beta}=\frac{25}{28}, \\
& \frac{\sin \beta \cos \alpha+\cos \beta \sin \alpha}{\sin \alpha \sin \beta}=\frac{25}{28} .
\end{aligned}
$$

Given that $\operatorname{ctg} \alpha=\frac{3}{4}, \frac{\cos \alpha}{\sin \alpha}=\frac{3}{4}, \frac{\cos ^{2} \alpha}{\sin ^{2} \alpha}=\frac{9}{16}, \frac{1-\sin ^{2} \alpha}{\sin ^{2} \alpha}=\frac{9}{16}$, we have

$$
\sin ^{2} \alpha=\frac{16}{25} \text { and, since } \alpha \in\left(0 ; \frac{\pi}{2}\right) \text {, then }
$$

$$
\sin \alpha=\frac{4}{5} ; \operatorname{ctg} \beta=\frac{1}{7}, \frac{\cos ^{2} \beta}{\sin ^{2} \beta}=\frac{1}{49}, \frac{1-\sin ^{2} \beta}{\sin ^{2} \beta}=\frac{1}{49}
$$

from which

$$
\sin ^{2} \beta=\frac{49}{50} \text { and, since } \beta \in\left(0 ; \frac{\pi}{2}\right) \text {, then }
$$

$$
\sin \beta=\frac{7}{\sqrt{50}}=\frac{7}{5 \sqrt{2}}
$$

Substituting the obtained values, we get

$$
\frac{\sin (\alpha+\beta)}{\frac{4}{5} \cdot \frac{7}{5 \sqrt{2}}}=\frac{25}{28}, \sin (\alpha+\beta)=\frac{\sqrt{2}}{2}
$$

from which $\alpha+\beta=(-1)^{n} \frac{\pi}{4}+\pi n, n \in Z$; and considering the constraints on $\alpha, \beta$, we have $\alpha+\beta=\frac{3 \pi}{4}$.

Answer: $\alpha+\beta=\frac{3 \pi}{4}$.