8.181. $\frac{1}{2 \operatorname{ctg}^{2} t+1}+\frac{1}{2 \operatorname{tg}^{2} t+1}=\frac{15 \cos 4 t}{8+\sin ^{2} 2 t}$.

8.181. $\frac{1}{2 \cot^{2} t+1}+\frac{1}{2 \tan^{2} t+1}=\frac{15 \cos 4 t}{8+\sin ^{2} 2 t}$.
## Solution:

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

Using the formulas

$$
\operatorname{ctg}^{2} \frac{\alpha}{2}=\frac{1+\cos \alpha}{1-\cos \alpha}, \operatorname{tg}^{2} \frac{\alpha}{2}=\frac{1-\cos \alpha}{1+\cos \alpha}, \cos 2 \alpha=2 \cos ^{2} \alpha-1
$$

$$
\begin{aligned}
& \frac{1}{\frac{2(1+\cos 2 t)}{1-\cos 2 t}+1}+\frac{1}{\frac{2(1-\cos 2 t)}{1+\cos 2 t}+1}=\frac{15\left(2 \cos ^{2} 2 t-1\right)}{8+1-\cos ^{2} 2 t} \cdot \Leftrightarrow \\
& \Leftrightarrow(1-\cos 2 t)(3-\cos 2 t)+(1+\cos 2 t)(3+\cos 2 t)=30 \cos ^{2} 2 t-15 \Leftrightarrow \\
& \Leftrightarrow 3-3 \cos 2 t-\cos 2 t+\cos ^{2} 2 t+3+3 \cos 2 t+\cos 2 t+ \\
& +\cos ^{2} 2 t-30 \cos ^{2} 2 t+15=0 \Leftrightarrow \\
& \Leftrightarrow 21-28 \cos ^{2} 2 t=0,4 \cos ^{2} 2 t=3, \cos ^{2} 2 t=\frac{3}{4}, \cos 2 t= \pm \frac{\sqrt{3}}{2} \\
& 2 t= \pm \frac{\pi}{6}+\pi k, t= \pm \frac{\pi}{12}+\frac{\pi}{2}=\frac{\pi}{12}(6 k \pm 1), \quad k \in Z \\
& \text { Answer: } t=\frac{\pi}{12}(6 k \pm 1), \quad k \in Z .
\end{aligned}
$$