$\frac{\tan \theta +\sec \theta -1}{\tan \theta -\sec \theta +1}=$

Solution:

$\frac{\tan \theta +\sec \theta -1}{\tan \theta -\sec \theta +1}=\frac{\sin \theta +1-\cos \theta }{\sin \theta -1+\cos \theta }$

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