#### Answer

The solution is
$$\lim_{x\to\pi/2^-}\left(\frac{\pi}{2}-x\right)\sec x=1.$$

#### Work Step by Step

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