3.176. Given: $\operatorname{ctg} \alpha=4, \operatorname{ctg} \beta=\frac{5}{3}, 0<\alpha<\frac{\pi}{2}, 0<\beta<\frac{\pi}{2}$. Find $\alpha+\beta$.
Solution.

$$
\operatorname{ctg} \alpha=4, \frac{\cos \alpha}{\sin \alpha}=4, \frac{\cos ^{2} \alpha}{\sin ^{2} \alpha}=16, \frac{1-\sin ^{2} \alpha}{\sin ^{2} \alpha}=16
$$

from which $\sin ^{2} \alpha=\frac{1}{17}$, hence for $0<\alpha<\frac{\pi}{2}$ we have

$$
\sin \alpha=\frac{1}{\sqrt{17}} ; \operatorname{ctg} \beta=\frac{\cos \beta}{\sin \beta}=\frac{5}{3}, \frac{\cos ^{2} \beta}{\sin ^{2} \beta}=\frac{25}{9}, \frac{1-\sin ^{2} \beta}{\sin ^{2} \beta}=\frac{25}{9}
$$

from which $\sin ^{2} \beta=\frac{9}{34}$, therefore, for $0<\beta<\frac{\pi}{2}$ we have $\sin \beta=\frac{3}{\sqrt{34}}$.

Then $\operatorname{ctg} \alpha+\operatorname{ctg} \beta=\frac{\cos \alpha}{\sin \alpha}+\frac{\cos \beta}{\sin \beta}=\frac{\sin \beta \cos \alpha+\sin \alpha \cos \beta}{\sin \alpha \sin \beta}=4+\frac{5}{3}=\frac{17}{3}$.

$\frac{\sin (\alpha+\beta)}{\sin \alpha \sin \beta}=\frac{17}{3}$.

Using the found values $\sin \alpha=\frac{1}{\sqrt{17}}$ and $\sin \beta=\frac{3}{\sqrt{34}}$, we have

$$
\frac{\sin (\alpha+\beta)}{\frac{1}{\sqrt{17}} \cdot \frac{3}{\sqrt{34}}}=\frac{17}{3}, \quad \frac{17 \sqrt{2} \sin (\alpha+\beta)}{3}=\frac{17}{3}
$$

from which $\sin (\alpha+\beta)=\frac{\sqrt{2}}{2}$.

Hence $\alpha+\beta=\frac{\pi}{4}$ for $0<\alpha<\frac{\pi}{2}$ and $0<\beta<\frac{\pi}{2}$.

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