3.261. $\frac{\sin \left(4 \alpha+\frac{5 \pi}{2}\right)}{1+\cos \left(4 \alpha-\frac{3 \pi}{2}\right)}$.

3.261. $\frac{\sin \left(4 \alpha+\frac{5 \pi}{2}\right)}{1+\cos \left(4 \alpha-\frac{3 \pi}{2}\right)}$. 

(Note: The mathematical expression is already in a universal format and does not require translation.)
## Solution.

$$
\begin{aligned}
& \frac{\sin \left(4 \alpha+\frac{5 \pi}{2}\right)}{1+\cos \left(4 \alpha-\frac{3 \pi}{2}\right)}=\frac{\sin \left(\frac{5 \pi}{2}+4 \alpha\right)}{1+\cos \left(\frac{3 \pi}{2}-4 \alpha\right)}=\frac{\cos 4 \alpha}{1-\sin 4 \alpha}=\frac{\cos 2(2 \alpha)}{1-\sin 2(2 \alpha)}= \\
&= \frac{\cos ^{2} 2 \alpha-\sin ^{2} 2 \alpha}{1-2 \sin 2 \alpha \cos 2 \alpha}=\frac{(\cos 2 \alpha-\sin 2 \alpha)(\cos 2 \alpha+\sin 2 \alpha)}{\cos ^{2} 2 \alpha-2 \sin 2 \alpha \cos 2 \alpha+\sin ^{2} 2 \alpha}= \\
&= \frac{(\cos 2 \alpha-\sin 2 \alpha)(\cos 2 \alpha+\sin 2 \alpha)}{(\cos 2 \alpha-\sin 2 \alpha)^{2}}=\frac{\cos 2 \alpha+\sin 2 \alpha}{\cos 2 \alpha-\sin 2 \alpha}= \\
&= \frac{1+\tan 2 \alpha}{1-\tan 2 \alpha}=\frac{\tan \frac{\pi}{4}+\tan 2 \alpha}{1-\tan \frac{\pi}{4} \tan 2 \alpha}= \\
&= \left[\tan(x+y)=\frac{\tan x+\tan y}{1-\tan x \tan y}, x, y, x+y \neq \frac{\pi}{2}+\pi n, n \in Z\right]= \\
&= \tan\left(\frac{\pi}{4}+2 \alpha\right) \\
& \text { Answer: } \tan\left(\frac{\pi}{4}+2 \alpha\right)
\end{aligned}
$$