8.317. $\frac{\sin ^{2} t-\tan^{2} t}{\cos ^{2} t-\cot^{2} t}+2 \tan^{3} t+1=0$.
## Solution.

Domain of definition: $\left\{\begin{array}{l}\cos t \neq 0, \\ \sin t \neq 0, \\ \sin t \neq \pm 1 .\end{array}\right.$

Rewrite the equation as

$\frac{\sin ^{2} t-\frac{\sin ^{2} t}{\cos ^{2} t}}{\cos ^{2} t-\frac{\cos ^{2} t}{\sin ^{2} t}}+2 \operatorname{tg}^{3} t+1=0 \Leftrightarrow$

$\Leftrightarrow \frac{\sin ^{2} t\left(\cos ^{2} t-1\right)}{\cos ^{2} t} \cdot \frac{\sin ^{2} t}{\cos ^{2} t\left(\sin ^{2} t-1\right)}+2 \operatorname{tg}^{3} t+1=0$,

$\left(\operatorname{tg}^{3} t\right)^{2}+2 \operatorname{tg}^{3} t+1=0,\left(\operatorname{tg}^{3} t+1\right)^{2}=0, \operatorname{tg}^{3} t=-1, \operatorname{tg} t=-1$,

$t=-\frac{\pi}{4}+\pi k=\frac{\pi}{4}(4 k-1), k \in Z$.

Answer: $t=\frac{\pi}{4}(4 k-1), k \in Z$.