8.225. $\operatorname{tg} t=\frac{\sin ^{2} t+\sin 2 t-1}{\cos ^{2} t-\sin 2 t+1}$.
## Solution.

Domain of definition: $\cos t \neq 0$.

Since $\sin ^{2} \frac{\alpha}{2}=\frac{1-\cos \alpha}{2}$ and $\cos ^{2} \frac{\alpha}{2}=\frac{1+\cos \alpha}{2}$, we have

$$
\begin{aligned}
& \operatorname{tg} t-\frac{\frac{1}{2}(1-\cos 2 t)+\sin 2 t-1}{\frac{1}{2}(1+\cos 2 t)-\sin 2 t+1}=0 \Leftrightarrow \operatorname{tg} t-\frac{1-\cos 2 t+2 \sin 2 t-2}{1+\cos 2 t-2 \sin 2 t+2}=0 \Leftrightarrow \\
& \Leftrightarrow \operatorname{tg} t-\frac{2 \sin 2 t-\cos 2 t-1}{-2 \sin 2 t+\cos 2 t+3}=0
\end{aligned}
$$

Applying the formulas $\sin \alpha=\frac{2 \operatorname{tg} \frac{\alpha}{2}}{1+\operatorname{tg}^{2} \frac{\alpha}{2}}$ and $\cos \alpha=\frac{1-\operatorname{tg}^{2} \frac{\alpha}{2}}{1+\operatorname{tg}^{2} \frac{\alpha}{2}}$, we get

$$
\begin{aligned}
& \operatorname{tg} t-\frac{\frac{4 \operatorname{tg} t}{1+\operatorname{tg}^{2} t}-\frac{1-\operatorname{tg}^{2} t}{1+\operatorname{tg}^{2} t}-1}{\frac{4 \operatorname{tg} t_{t}}{1+\operatorname{tg}^{2} t}+\frac{1-\operatorname{tg}^{2} t}{1+\operatorname{tg}^{2} t}+3}=0 \Leftrightarrow \operatorname{tg}^{3} t-2 \operatorname{tg}^{2} t+1=0 \Leftrightarrow \\
& \Leftrightarrow\left(\operatorname{tg}^{3} t-\operatorname{tg}^{2} t\right)-\left(\operatorname{tg}^{2} t-1\right)=0 \Leftrightarrow \operatorname{tg}^{2} t(\operatorname{tg} t-1)-(\operatorname{tg} t-1)(\operatorname{tg} t+1)=0 \Leftrightarrow \\
& \Leftrightarrow(\operatorname{tg} t-1)\left(\operatorname{tg}^{2} t-\operatorname{tg} t-1\right)=0
\end{aligned}
$$

From this, we have either $\operatorname{tg} t-1=0, \operatorname{tg} t=1, t_{1}=\frac{\pi}{4}+\pi k=\frac{\pi}{4}(4 k+1), k \in Z$, or $\operatorname{tg}^{2} t-\operatorname{tg} t-1=0$. Solving this equation as a quadratic in $\operatorname{tg} t$, we get $\operatorname{tg} t=\frac{1-\sqrt{5}}{2}, t_{2}=\operatorname{arctg} \frac{1-\sqrt{5}}{2}+\pi n, n \in Z ; \operatorname{tg} t=\frac{1+\sqrt{5}}{2}$, $t_{3}=\operatorname{arctg} \frac{1+\sqrt{5}}{2}+\pi l, l \in Z$.

Answer: $t_{1}=\frac{\pi}{4}(4 k+1) ; t_{2}=\operatorname{arctg} \frac{1-\sqrt{5}}{2}+\pi n ; t_{3}=\operatorname{arctg} \frac{1+\sqrt{5}}{2}+\pi l ; k$, $n$ and $l \in Z$.