## Precalculus (10th Edition)

$-2$
RECALL: (1) $\sec{\theta}=\frac{1}{\cos{\theta}}$ (2) $\csc{\theta} = \frac{1}{\sin{\theta}}$ With $\sin{\frac{\pi}{2}}=1$ and $\cos{\pi}=-1$, then $\sec\pi -csc{\frac {\pi }{2}} \\=\dfrac {1}{\cos \pi }-\dfrac {1}{\sin{\frac {\pi }{2}}} \\=\dfrac {1}{-1}-\dfrac {1}{1} \\=-1-1 \\=-2$