3.301. $\frac{2 \cos ^{2} 2 \alpha-1}{2 \operatorname{tg}\left(\frac{\pi}{4}-2 \alpha\right) \sin ^{2}\left(\frac{3}{4} \pi-2 \alpha\right)}-\operatorname{tg} 2 \alpha+\cos 2 \alpha-\sin 2 \alpha$.
## Solution.

$$
\begin{aligned}
& \frac{2 \cos ^{2} 2 \alpha-1}{2 \operatorname{tg}\left(\frac{\pi}{4}-2 \alpha\right) \sin ^{2}\left(\frac{3}{4} \pi-2 \alpha\right)}-\operatorname{tg} 2 \alpha+\cos 2 \alpha-\sin 2 \alpha= \\
& =\left[2 \cos ^{2} 2 x-1=\cos 4 x ; \operatorname{tg} \frac{x}{2}=\frac{\sin x}{1+\cos x}\right. \\
& \left.x \neq \pi+2 \pi n, n \in Z, \sin ^{2} \frac{x}{2}=\frac{1-\cos x}{2}\right]= \\
& =\frac{\cos 4 \alpha}{2 \cdot \frac{\sin \left(\frac{\pi}{2}-4 \alpha\right)}{1+\cos \left(\frac{\pi}{2}-4 \alpha\right)} \cdot \frac{1-\cos \left(\frac{3 \pi}{2}-4 \alpha\right)}{2}}-\operatorname{tg} 2 \alpha+\cos 2 \alpha-\sin 2 \alpha= \\
& =\frac{\cos 4 \alpha}{\frac{\cos 4 \alpha}{1+\sin 4 \alpha} \cdot(1+\sin 4 \alpha)}-\operatorname{tg} 2 \alpha+\cos 2 \alpha-\sin 2 \alpha=
\end{aligned}
$$

$$
\begin{aligned}
& =1-\operatorname{tg} 2 \alpha+\cos 2 \alpha-\sin 2 \alpha=\left(1-\frac{\sin 2 \alpha}{\cos 2 \alpha}\right)+(\cos 2 \alpha-\sin 2 \alpha)= \\
& =\frac{\cos 2 \alpha-\sin 2 \alpha}{\cos 2 \alpha}+(\cos 2 \alpha-\sin 2 \alpha)=(\cos 2 \alpha-\sin 2 \alpha)\left(\frac{1}{\cos 2 \alpha}+1\right)= \\
& =\frac{(\cos 2 \alpha-\sin 2 \alpha)(1+\cos 2 \alpha)}{\cos 2 \alpha}= \\
& =\frac{\frac{2}{\sqrt{2}}\left(\frac{\sqrt{2}}{2} \cos 2 \alpha-\frac{\sqrt{2}}{2} \sin 2 \alpha\right)\left(1+\cos ^{2} \alpha-\sin ^{2} \alpha\right)}{\cos 2 \alpha}= \\
& =\frac{2 \sqrt{2}\left(\sin \frac{\pi}{4} \cos 2 \alpha-\cos \frac{\pi}{4} \sin 2 \alpha\right) \cos ^{2} \alpha}{\cos 2 \alpha}= \\
& =[\sin x \cos y-\cos x \sin y=\sin (x-y)]=\frac{2 \sqrt{2} \sin \left(\frac{\pi}{4}-2 \alpha\right) \cos ^{2} \alpha}{\cos 2 \alpha} \\
& \text { Answer: } \frac{2 \sqrt{2} \sin \left(\frac{\pi}{4}-2 \alpha\right) \cos ^{2} \alpha}{\cos 2 \alpha} . \\
& \text { 3.302. } \frac{\operatorname{ctg} 2 \alpha+\operatorname{tg} 2 \alpha}{1+\operatorname{tg} 4 \alpha \operatorname{tg} 2 \alpha} .
\end{aligned}
$$

Solution.

$$
\begin{aligned}
& \frac{\operatorname{ctg} 2 \alpha+\operatorname{tg} 2 \alpha}{1+\operatorname{tg} 4 \alpha \operatorname{tg} 2 \alpha}=\frac{\frac{\cos 2 \alpha}{\sin 2 \alpha}+\frac{\sin 2 \alpha}{\cos 2 \alpha}}{1+\frac{\sin 4 \alpha \sin 2 \alpha}{\cos 4 \alpha \cos 2 \alpha}}=\frac{\frac{\cos ^{2} 2 \alpha+\sin ^{2} 2 \alpha}{\sin 2 \alpha \cos 2 \alpha}}{\frac{\cos 4 \alpha \cos 2 \alpha+\sin 4 \alpha \sin 2 \alpha}{\cos 4 \alpha \cos 2 \alpha}}= \\
& =\frac{\cos ^{2} 2 \alpha+\sin ^{2} 2 \alpha}{\sin 2 \alpha \cos \alpha} \cdot \frac{\cos 4 \alpha \cos 2 \alpha}{\cos 4 \alpha \cos 2 \alpha+\sin 4 \alpha \sin 2 \alpha}= \\
& =\frac{\cos 4 \alpha}{\sin 2 \alpha(\cos 4 \alpha \cos 2 \alpha+\sin 4 \alpha \sin 2 \alpha)}=
\end{aligned}
$$

$=[\cos x \cos y+\sin x \sin y=\cos (x-y)]=\frac{\cos 4 \alpha}{\sin 2 \alpha \cos 2 \alpha}=\frac{2 \cos +\alpha}{2 \sin 2 \alpha \cos 2 \alpha}$

$=\frac{2 \cos 4 \alpha}{\sin 4 \alpha}=2 \operatorname{ctg} 4 \alpha$.

Answer: $2 \operatorname{ctg} 4 \alpha$.