If $\mathrm{x}=\frac{2\sin \mathrm{\theta }}{1+\cos \mathrm{\theta }+\sin \mathrm{\theta }}, \mathrm{then} \frac{1-\cos \mathrm{\theta }+\sin \mathrm{\theta }}{1+\sin \mathrm{\theta }}$ is equal to ______

$$\begin{gathered} x=\frac{2\sin \theta }{1+\cos \theta +\sin \theta } \\ x=\left(\frac{2\sin \theta }{1+\sin \theta +\cos \theta }\right)\left(\frac{1+\sin \theta -\cos \theta }{1+\sin \theta -\cos \theta }\right) \\ x=\frac{2\sin \theta \left(1+\sin \theta -\cos \theta \right)}{{\left(1+\sin \theta \right)}^2-{\cos }^2\theta }=\frac{2\sin \theta \left(1+\sin \theta -\cos \theta \right)}{2\sin \theta \left(1+\sin \theta \right)} \\ x=\frac{1+\sin \theta -\cos \theta }{1+\sin \theta } \end{gathered}$$