We have $\frac{2\sin \alpha }{1+\cos \alpha +\sin \alpha }=x.$

$\Rightarrow \frac{2\sin \alpha }{1+\cos \alpha +\sin \alpha }\times \frac{1-\left(\cos \alpha +\sin \alpha \right)}{1-\left(\cos \alpha +\sin \alpha \right)}=x$

$\Rightarrow \frac{2\sin \alpha \left(1-\cos \alpha -\sin \alpha \right)}{1^2-{\left(\cos \alpha +\sin \alpha \right)}^2}=x$

$\Rightarrow \frac{2\sin \alpha \left(1-\cos \alpha -\sin \alpha \right)}{1-\left({\cos }^2\alpha +{\sin }^2\alpha +2\cos \alpha \sin \alpha \right)}=x$

Using ${\sin }^2\theta +{\cos }^2\theta =1,$

$\Rightarrow \frac{2\sin \alpha \left(1-\cos \alpha -\sin \alpha \right)}{1-\left(1+2\cos \alpha \sin \alpha \right)}=x$

$\Rightarrow \frac{2\sin \alpha \left(1-\cos \alpha -\sin \alpha \right)}{1-1-2\cos \alpha \sin \alpha }=x$

$\Rightarrow \frac{2\sin \alpha \left(1-\cos \alpha -\sin \alpha \right)}{-2\cos \alpha \sin \alpha }=x$

$\Rightarrow \frac{\left(1-\cos \alpha -\sin \alpha \right)}{-\cos \alpha }=x$

$\Rightarrow \frac{\left(1-\cos \alpha -\sin \alpha \right)}{\cos \alpha }=-x.$