Answer
$$0$$
Work Step by Step
We have
$$
\lim _{x \rightarrow \pi/2}\left(\sec x- \tan x\right)=\lim _{x \rightarrow \pi/2}\left(\frac{1}{\cos x}- \frac{\sin x}{\cos x}\right)\\
=\lim _{x \rightarrow \pi/2}\left(\frac{1-\sin x}{\cos x}\right)=\frac{0}{0}
$$
is an intermediate form, so we can apply L’Hôpital’s Rule as follows
$$\lim _{x \rightarrow \pi/2}\left(\frac{1-\sin x}{\cos x}\right)=\lim _{x \rightarrow \pi/2}\left(\frac{-\cos x}{\sin x}\right)=\frac{0}{1}=0.$$
