Express the complex number $z=\frac{1+\sin \theta+\mathrm{i} \cos \theta}{1+\sin \theta-\mathrm{i} \cos \theta}$ in polar form as $\qquad$
10. $\cos \left(\frac{\pi}{2}-\theta\right)+i \sin \left(\frac{\pi}{2}-\theta\right)$.

Given $\sin ^{2} \theta-\cos ^{2} \theta=1$, we have $(\sin \theta+\mathrm{i} \cos \theta)(\sin \theta-\mathrm{i} \cos \theta)=1$. Converting it into a ratio form, we get $\frac{\sin \theta+\mathrm{i} \cos \theta}{1}=\frac{1}{\sin \theta-i \cos \theta}=\frac{1-\sin \theta+\mathrm{i} \cos \theta}{1+\sin \theta-\mathrm{i} \cos \theta}$. Therefore, $z-\sin \theta+\mathrm{i} \cos \theta=\cos \left(\frac{\pi}{2}-\theta\right)+\mathrm{i} \sin \left(\frac{\pi}{2}-\theta\right)$.