#### Answer

$ -\infty$

#### Work Step by Step

$x\rightarrow(\pi/2)^{+}$ means that
x approaches $\displaystyle \frac{\pi}{2}$ from the right (the greater side),
it is positive, so $\displaystyle \frac{1}{x}$ is positive.
Secant and cosine are reciprocal.
The cosine (and secant) of x, belonging (slightly) to quadrant II, are negative.
Since $\cos x$ approaches 0 from the negative side, $\sec x\rightarrow-\infty$
So, as $x\rightarrow(\pi/2)^{+}$
the product ($\displaystyle \frac{1}{x}\sec x$) $\rightarrow-\infty$.