The Polar form of $1 + \mathrm{itan\theta }\left(-\mathrm{\pi } <\hspace{0.17em}\mathrm{\theta } < -\frac{\mathrm{\pi }}{2}\right)$ is

$$\begin{gathered} -\mathrm{\pi }<\mathrm{\theta }<\frac{-\mathrm{\pi }}{2}\Rightarrow 0<\mathrm{\pi }+\mathrm{\theta }<\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$2$}\right. \\ 1+\mathrm{i} \tan \mathrm{\theta }=1+\mathrm{i} \frac{\mathrm{sin\theta }}{\mathrm{cos\theta }} \\ =\frac{1}{\mathrm{cos\theta }}(\mathrm{cos\theta }+\mathrm{i} \mathrm{sin\theta }) \\ =-\mathrm{sec\theta }(-\mathrm{cos\theta }-\mathrm{i} \mathrm{sin\theta }) \\ =-\mathrm{sec\theta }\left(\cos \right(\mathrm{\pi }+\mathrm{\theta })+\mathrm{i} \sin (\mathrm{\pi }+\mathrm{\theta }\left)\right) \\ =-\mathrm{sec\theta } \mathrm{cis} (\mathrm{\pi }+\mathrm{\theta }) \end{gathered}$$