Limit: $\lim\limits_{x\to{\pi/2}^+} \frac{\ln(x-\pi/2)}{\tan(x)}$ using De l'Hôpital's rule?

Find the limit

$$\lim_{x \to {\pi/2}^+} \frac{\ln(x-\pi/2)}{\tan(x)}$$

So, both the numerator and the denominator approach negative infinity. There De l'Hôpital's rule applies. I found the derivative of both and got:

$$\lim_{x \to {\pi/2}^+} \frac{{1/(x-\pi/2)}}{\sec^2(x)}$$

I keep on applying De l'Hôpital's rule yet still cannot find an expression that will not result in zero as the denominator.