$3.455 A=\operatorname{tg}\left(2 \arccos \frac{5}{\sqrt{26}}-\arcsin \frac{12}{13}\right)$. 

$3.455 A=\tan\left(2 \arccos \frac{5}{\sqrt{26}}-\arcsin \frac{12}{13}\right)$.
## Solution.

Let $\alpha=\arccos \frac{5}{\sqrt{26}}, \beta=\arcsin \frac{12}{13} \Rightarrow \cos \alpha=\frac{5}{\sqrt{26}}, \operatorname{tg} \alpha=\frac{1}{5}$,

$\sin \beta=\frac{12}{13}, \operatorname{tg} \beta=\frac{12}{5} \Rightarrow A=\operatorname{tg}(2 \alpha-\beta)=\frac{\operatorname{tg} 2 \alpha-\operatorname{tg} \beta}{1+\operatorname{tg} 2 \alpha \cdot \operatorname{tg} \beta}=\frac{\frac{2 \operatorname{tg} \alpha}{1-\operatorname{tg}^{2} \alpha}-\operatorname{tg} \beta}{1+\frac{2 \operatorname{tg} \alpha \cdot \operatorname{tg} \beta}{1-\operatorname{tg}^{2} \alpha}}=-\frac{119}{120}$.

Answer: $A=-\frac{119}{120}$.