Answer
The solution is
$$\lim_{\theta\to\pi/2^-}(\tan\theta-\sec\theta)=0.$$
Work Step by Step
To solve this limit we will use L'Hopital's rule. "LR" will stand for "Apply L'Hopital's rule".
$$\lim_{\theta\to\pi/2^-}(\tan\theta-\sec\theta)=\lim_{\theta\to\pi/2^-}\left(\frac{\sin\theta}{\cos\theta}-\frac{1}{\cos \theta}\right)=\lim_{\theta\to\pi/2^-}\frac{\sin\theta-1}{\cos\theta}=\left[\frac{\sin\frac{\pi}{2}^--1}{\cos\pi/2^-}\right]=\left[\frac{0}{0}\right][\text{LR}]=\lim_{\theta\to\pi/2^-}\frac{(\sin\theta-1)'}{(\cos\theta)'}=\lim_{\theta\to\pi/2^-}\frac{\cos\theta}{-\sin\theta}=\lim_{\theta\to\pi/2^-}-\cot x=\left[-\cot\frac{\pi}{2}^-\right]=0.$$
