$3.462 A=\operatorname{tg}\left(\frac{1}{2} \arccos \frac{3}{5}-2 \operatorname{arctg}(-2)\right)$

$3.462 A=\tan\left(\frac{1}{2} \arccos \frac{3}{5}-2 \arctan(-2)\right)$
Solution. Since $\operatorname{arctg}(-2)=\operatorname{arcctg}\left(-\frac{1}{2}\right)-\pi$, this problem is analogous to problem 3.457.

Answer: $A=-\frac{1}{2}$.