$(\tan\theta + \sec\theta)^2 = ?$

Expanding, $(\tan \theta +\sec \theta {)}^2={\left({\displaystyle \frac{\sin \theta +1}{\cos \theta }}\right)}^2={\displaystyle \frac{(1+\sin \theta {)}^2}{{\cos }^2\theta }}={\displaystyle \frac{(1+\sin \theta {)}^2}{1-{\sin }^2\theta }}={\displaystyle \frac{(1+\sin \theta {)}^2}{(1+\sin \theta )(1-\sin \theta )}}={\displaystyle \frac{1+\sin \theta }{1-\sin \theta }}$.