$3.458 A=\operatorname{tg}\left(\frac{1}{2} \arccos \frac{3}{5}-3 \operatorname{arcctg}(-2)\right)$

$3.458 A=\tan\left(\frac{1}{2} \arccos \frac{3}{5}-3 \operatorname{arcctg}(-2)\right)$
## Solution.

Let $\alpha=\arccos \frac{3}{5}, \beta=\operatorname{arcctg}(-2) \Rightarrow \cos \alpha=\frac{3}{5}, \sin \alpha=\frac{4}{5}$,

$\operatorname{tg} \frac{\alpha}{2}=\frac{\sin \alpha}{1+\cos \alpha}=\frac{1}{2}, \operatorname{ctg} \beta=-2, \operatorname{tg} \beta=-\frac{1}{2}, \operatorname{tg} 3 \beta=\frac{3 \operatorname{tg} \beta-\operatorname{tg}^{3} \beta}{1-3 \operatorname{tg}^{2} \beta}=-\frac{11}{2} \Rightarrow$

$\Rightarrow A=\operatorname{tg}\left(\frac{\alpha}{2}-3 \beta\right)=\frac{\operatorname{tg} \frac{\alpha}{2}-\operatorname{tg} 3 \beta}{1+\operatorname{tg} \frac{\alpha}{2} \cdot \operatorname{tg} 3 \beta}=-\frac{24}{7}$.

Answer: $A=-\frac{24}{7}$.