I'm trying to compute the following limit and would greatly appreciate your heartening feedback on my solution.

The limit:

$\lim_{\theta \to \frac{\pi}{2}} (\sec \theta - \tan \theta)$

My steps in deriving the solution:

Preliminary identities:

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

$\frac{1}{\cos \theta}-\frac{\sin\theta}{\cos\theta} = \frac{1-\sin\theta}{\cos\theta} = \frac{1-\sin^2\theta}{(1+\sin\theta)\cos\theta} = \frac{\cos^2\theta}{\cos\theta}\cdot \frac{1}{1+\sin\theta} = \frac{\cos\theta}{1+\sin\theta} = \frac{0}{1+1}$

When $\theta \to \frac{\pi}{2}$ then $\cos(\frac{\pi}{2}) = 0$ and $\sin(\frac{\pi}{2}) = 1$

The answer being:

$\lim_{\theta \to \frac{\pi}{2}} (\sec \theta - \tan \theta) = 0$