8.257. $\operatorname{tg}^{2} \frac{z}{2}+\operatorname{ctg}^{2} \frac{z}{2}-2=4 \operatorname{tg} z$.
## Solution.

Domain of definition: $\left\{\begin{array}{l}\cos z \neq 0, \\ \sin z \neq 0 .\end{array}\right.$

Write the equation in the form

$$
\begin{aligned}
& \left(\operatorname{tg} \frac{z}{2}+\operatorname{ctg} \frac{z}{2}\right)^{2}-2 \operatorname{tg} \frac{z}{2} \operatorname{ctg} \frac{z}{2}-2-4 \operatorname{tg} z=0 \Leftrightarrow \\
& \Leftrightarrow\left(\frac{\sin \frac{z}{2}}{\cos \frac{z}{2}}+\frac{\cos \frac{z}{2}}{\sin \frac{z}{2}}\right)^{2}-4-\frac{4 \sin z}{\cos z}=0 \Leftrightarrow \\
& \Leftrightarrow\left(\frac{\sin ^{2} \frac{z}{2}+\cos ^{2} \frac{z}{2}}{\sin \frac{z}{2} \cos \frac{z}{2}}\right)^{2}-4-\frac{4 \sin z}{\cos z}=0 \Leftrightarrow \frac{4}{\sin ^{2} z}-4-\frac{4 \sin z}{\cos z}=0 \Leftrightarrow \\
& \Leftrightarrow \cos z-\sin ^{2} z \cos z-\sin ^{3} z=0 \Leftrightarrow \cos z\left(1-\sin ^{2} z\right)-\sin ^{3} z=0 \Leftrightarrow \\
& \Leftrightarrow \cos z-\sin ^{3} z=0 \Leftrightarrow(\cos z-\sin z)\left(\cos ^{2} z+\sin z \cos z+\sin ^{2} z\right)=0 \Leftrightarrow \\
& \Leftrightarrow\left(\cos z-\sin ^{2} z\right)(1+\sin z \cos z)=0 .
\end{aligned}
$$

## From this

1) $\cos z-\sin z=0, \operatorname{tg} z=1, z_{1}=\frac{\pi}{4}+\pi n=\frac{\pi}{4}(4 n+1), n \in Z$;
2) $1+\sin z \cos z=0, \frac{1}{2} \sin 2 z=-1, \sin 2 z=-2, \varnothing$.

Answer: $z=\frac{\pi}{4}(4 n+1), n \in Z$.