Answer
$-1$
Work Step by Step
Consider: $\lim\limits_{x \to (\dfrac{\pi}{2})^{-}}f(x)=\lim\limits_{x \to (\dfrac{\pi}{2})^{-}} (x-\dfrac{\pi}{2}) \sec x$
We need to check that the limit has an indeterminate form.
Thus, $f((\dfrac{\pi}{2})^{-})=\dfrac{0}{0}$
The limit shows an indeterminate form. Thus, apply L-Hospital's rule: $\lim\limits_{a \to b}f(x)=\lim\limits_{a \to b}\dfrac{g'(x)}{h'(x)}$
Then
$\lim\limits_{x \to (\dfrac{\pi}{2})^{-}}\dfrac{1}{-\sin x}=\dfrac{1}{-\sin (\pi/2)}=-1$
