Evaluating arcsec at infinity...?

**Nervous** · January 10th 2013, 05:33 AM

$arcsec(2) - arcsec(1) + arcsec(\infty) - arcsec(2)$
No idea how to procede, I see the first and last terms cancel out, my book says arcsec of 1 comes out to zero and infinity comes out to pi-halfs. How?

**ebaines** · January 10th 2013, 07:37 AM

The arcsecant of 1 is 0 because sec(0) = $\frac 1 {\cos(0)} = \frac 1 1 = 1$ .

The arcsecant of $\infty$ is $\frac {\pi} 2$ because $sec(\pi /2) = \frac 1 {\cos (\pi/2)} = \frac 1 0 = \infty$

**HallsofIvy** · January 10th 2013, 08:13 AM

Just to be pedantic- what ebaines really means $\sec(\pi/2)= \lim_{x\to \pi/2}\frac{1}{cos(x)}= \frac{1}{\lim_{x\to\pi/2} cos(x)}= \infty$ .

(I just refuse to say that " $\frac{1}{0}= \infty$ "!)

**ebaines** · January 10th 2013, 08:19 AM

Go ahead - be pedantic! It keeps us honest.

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