1. Simplify $\frac{(\cos \theta-\mathrm{i} \sin \theta)^{8} \cdot(1+\mathrm{i}\tan \theta)^{5}}{(\cos \theta+\mathrm{i} \sin \theta)^{2} \cdot(\tan \theta+\mathrm{i})}$.
\begin{array}{l}\text { 1. Since } 1+i \tan \theta=\frac{1}{\cos \theta}(\cos \theta+i \sin \theta) \text {, } \\ \tan \theta+i=\frac{\sin \theta}{\cos \theta}+i=\frac{1}{\cos \theta}(\sin \theta+i \cos \theta)=\frac{1}{\cos \theta}\left[\cos \left(\frac{\pi}{2}-\theta\right)+i\sin\left(\frac{\pi}{2}-\theta\right)\right] . \\ \text { Therefore, the original expression }=\frac{[\cos (-\theta)+i \sin (-\theta)]^{8} \cdot \frac{1}{\cos ^{5} \theta}(\cos \theta+i \sin \theta)^{5}}{(\cos \theta+i \sin \theta)^{2} \cdot \frac{1}{\cos \theta}\left[\cos \left(\frac{\pi}{2}-\theta\right)+i \sin \left(\frac{\pi}{2}-\theta\right)\right]} \\ =\frac{1}{\cos ^{4} \theta}\left[\cos \left(-4 \theta-\frac{\pi}{2}\right)+i \sin \left(-4 \theta-\frac{\pi}{2}\right)\right] \\ =\frac{1}{\cos ^{4} \theta}(-\sin 4 \theta-i \cos 4 \theta)=-\frac{1}{\cos ^{4} \theta}(\sin 4 \theta+i \cos 4 \theta) . \\\end{array}