If $0<\theta <\frac{\pi }{2},andif\frac{y+1}{1-y}=\sqrt{\frac{1+\sin \theta }{1-\sin \theta }},thenyisequalto$

$\frac{y+1}{1-y}=\frac{\sqrt{1+\sin \theta }}{1-\sin \theta }$

$\frac{y+1}{1-y}=\frac{\sqrt{{\left(\cos \frac{\theta }{2}+\sin \frac{\theta }{2}\right)}^2}}{{\left(\cos \frac{\theta }{2}-\sin \frac{\theta }{2}\right)}^2}$

$\frac{y+1}{1-y}=\frac{\cos \frac{\theta }{2}+\sin \frac{\theta }{2}}{\left|\cos \frac{\theta }{2}-\sin \frac{\theta }{2}\right|}$

$\frac{y+1}{1-y}=\frac{\cos \frac{\theta }{2}+\sin \frac{\theta }{2}}{\cos \frac{\theta }{2}-\sin \frac{\theta }{2}}$          $\left[\begin{array}{c}\because 0<\theta <\frac{\pi }{2}\Rightarrow 0<\frac{\theta }{2}<\frac{\pi }{4}\\ \Rightarrow \cos \frac{\theta }{2}>\sin \frac{\theta }{2}\end{array}\right]$

$\frac{y+1}{1-y}=\frac{1+\tan \frac{\theta }{2}}{1-\tan \frac{\theta }{2}}\Rightarrow y=\tan \frac{\theta }{2}$