3.284. $\operatorname{tg}\left(2 \operatorname{arctg}\left(\frac{1-\cos x}{\sin x}\right)\right) \sqrt{\frac{1+\cos 2 x}{1-\cos 2 x}}$.

3.284. $\tan\left(2 \arctan\left(\frac{1-\cos x}{\sin x}\right)\right) \sqrt{\frac{1+\cos 2 x}{1-\cos 2 x}}$.
## Solution.

$$
\begin{aligned}
& \operatorname{tg}\left(2 \operatorname{arctg}\left(\frac{1-\cos x}{\sin x}\right)\right) \sqrt{\frac{1+\cos 2 x}{1-\cos 2 x}}= \\
& =\frac{2 \operatorname{tg}\left(\operatorname{arctg} \frac{1-\cos x}{\sin x}\right)}{1-\operatorname{tg}^{2}\left(\operatorname{arctg} \frac{1-\cos x}{\sin x}\right)} \cdot \sqrt{\frac{1+\left(2 \cos ^{2} x-1\right)}{1-\left(1-2 \sin ^{2} x\right)}}= \\
& =\frac{2 \operatorname{tg}\left(\operatorname{arctg} \frac{1-\cos x}{\sin x}\right)}{1-\operatorname{tg}^{2}\left(\operatorname{arctg} \frac{1-\cos x}{\sin x}\right)} \cdot \sqrt{\frac{\cos ^{2} x}{\sin ^{2} x}}=\frac{2 \operatorname{tg}\left(\operatorname{arctg} \frac{1-\cos x}{\sin x}\right)}{1-\operatorname{tg}^{2}\left(\operatorname{arctg} \frac{1-\cos x}{\sin x}\right)} \cdot \sqrt{\operatorname{ctg}^{2} x}=
\end{aligned}
$$

$$
\begin{aligned}
& =\frac{2 \operatorname{tg}\left(\operatorname{arctg} \frac{1-\cos x}{\sin x}\right)}{1-\operatorname{tg}^{2}\left(\operatorname{arctg} \frac{1-\cos x}{\sin x}\right)} \cdot|\operatorname{ctg} x|=\frac{2 \cdot \frac{1-\cos x}{\sin x}}{1-\left(\frac{1-\cos x}{\sin x}\right)^{2}} \cdot|\operatorname{ctg} x|= \\
& =\frac{2(1-\cos x) \cdot|\operatorname{ctg} x|}{\sin x} \cdot \frac{\sin ^{2} x}{\sin ^{2} x-(1-\cos x)^{2}}=\frac{2(1-\cos x) \cdot|\operatorname{ctg} x| \sin x}{\left(1-\cos ^{2} x\right)-(1-\cos x)^{2}}= \\
& =\frac{2(1-\cos x) \cdot|\operatorname{ctg} x| \sin x}{(1-\cos x)(1+\cos x)-(1-\cos x)^{2}}=\frac{2(1-\cos x) \cdot|\operatorname{ctg} x| \sin x}{(1-\cos x)(1+\cos x-1+\cos x)}= \\
& =\frac{2|\operatorname{ctg} x| \sin x}{2 \cos x}=|\operatorname{ctg} x| \cdot \operatorname{tg} x=\left\{\begin{array}{l}
-\operatorname{ctg} x \operatorname{tg} x=-1, \text { if } \operatorname{ctg} x < 0 \\
\operatorname{ctg} x \operatorname{tg} x=1, \text { if } \operatorname{ctg} x > 0
\end{array}\right.
\end{aligned}
$$

Answer: 1, if $\operatorname{ctg} x > 0$; -1, if $\operatorname{ctg} x < 0$.

Transform into a product (3.285.-3.331):