3.465 Given that $\operatorname{tg} \alpha=m$, find

$A=\sin ^{2}\left(\frac{\pi}{4}+\alpha\right)-\sin ^{2}\left(\frac{\pi}{6}-\alpha\right)-\cos \frac{5 \pi}{12} \cdot \sin \left(\frac{5 \pi}{12}-2 \alpha\right)$
Solution.

$$
\begin{aligned}
& A=\frac{1-\cos \left(\frac{\pi}{2}+2 \alpha\right)}{2}-\frac{1-\cos \left(\frac{\pi}{3}-2 \alpha\right)}{2}-\frac{-\sin 2 \alpha+\sin \left(\frac{5 \pi}{6}-2 \alpha\right)}{2}= \\
& =\frac{1}{2}\left[2 \sin 2 \alpha+\cos \left(\frac{\pi}{3}-2 \alpha\right)-\sin \left(\frac{5 \pi}{6}-2 \alpha\right)\right]=\sin 2 \alpha\left(\text { since } \cos \left(\frac{\pi}{3}-2 \alpha\right)=\right. \\
& \left.=\cos \left(\frac{5 \pi}{6}-2 \alpha-\frac{\pi}{2}\right)=\sin \left(\frac{5 \pi}{6}-2 \alpha\right)\right) \Rightarrow A=\frac{2 \operatorname{tg} \alpha}{1+\operatorname{tg}^{2} \alpha}=\frac{2 m}{1+m^{2}} .
\end{aligned}
$$

Answer: $A=\frac{2 m}{1+m^{2}}$.