3.079. $\frac{1+\operatorname{cot} 2 \alpha \operatorname{cot} \alpha}{\operatorname{tan} \alpha+\operatorname{cot} \alpha}$.
## Solution.

$\frac{1+\operatorname{ctg} 2 \alpha \operatorname{ctg} \alpha}{\operatorname{tg} \alpha+\operatorname{ctg} \alpha}=\frac{1+\frac{\operatorname{ctg}^{2} \alpha-1}{2 \operatorname{ctg} \alpha} \cdot \operatorname{ctg} \alpha}{\frac{1}{\operatorname{ctg} \alpha}+\operatorname{ctg} \alpha}=\frac{\frac{2+\operatorname{ctg}^{2} \alpha-1}{2}}{\frac{1+\operatorname{ctg}^{2} \alpha}{\operatorname{ctg} \alpha}}=\frac{1+\operatorname{ctg}^{2} \alpha}{2} \times$

$\times \frac{\operatorname{ctg} \alpha}{1+\operatorname{ctg}^{2} \alpha}=\frac{\operatorname{ctg} \alpha}{2}$.

Answer: $\frac{\operatorname{ctg} \alpha}{2}$.