$3.457 A=\operatorname{ctg}\left(\frac{1}{2} \arccos \frac{3}{5}-2 \operatorname{arcctg}\left(-\frac{1}{2}\right)\right)$

$3.457 A=\cot\left(\frac{1}{2} \arccos \frac{3}{5}-2 \operatorname{arcctg}\left(-\frac{1}{2}\right)\right)$
Solution. Let $\alpha=\arccos \frac{3}{5}, \beta=\operatorname{arcctg}\left(-\frac{1}{2}\right) \Rightarrow \cos \alpha=\frac{3}{5}, \sin \alpha=\frac{4}{5}$, $\operatorname{ctg} \frac{\alpha}{2}=\frac{1+\cos \alpha}{\sin \alpha}=2, \operatorname{ctg} \beta=-\frac{1}{2}, \operatorname{ctg} 2 \beta=\frac{\operatorname{ctg}^{2} \beta-1}{2 \operatorname{ctg} \beta}=\frac{3}{4} \Rightarrow A=\operatorname{ctg}\left(\frac{\alpha}{2}-2 \beta\right)=$

$$
=\frac{\operatorname{ctg} \frac{\alpha}{2} \cdot \operatorname{ctg} 2 \beta+1}{-\operatorname{ctg} \frac{\alpha}{2}+\operatorname{ctg} 2 \beta}=-2
$$

Answer: $A=-2$.