$\displaystyle arcsec(2) - arcsec(1) + arcsec(\infty) - arcsec(2)$

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?

Printable View

- Jan 10th 2013, 05:33 AMNervousEvaluating arcsec at infinity...?
$\displaystyle arcsec(2) - arcsec(1) + arcsec(\infty) - arcsec(2)$

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? - Jan 10th 2013, 07:37 AMebainesRe: Evaluating arcsec at infinity...?
The arcsecant of 1 is 0 because sec(0) = $\displaystyle \frac 1 {\cos(0)} = \frac 1 1 = 1$.

The arcsecant of $\displaystyle \infty$ is $\displaystyle \frac {\pi} 2$ because $\displaystyle sec(\pi /2) = \frac 1 {\cos (\pi/2)} = \frac 1 0 = \infty$ - Jan 10th 2013, 08:13 AMHallsofIvyRe: Evaluating arcsec at infinity...?
Just to be pedantic- what ebaines really means $\displaystyle \sec(\pi/2)= \lim_{x\to \pi/2}\frac{1}{cos(x)}= \frac{1}{\lim_{x\to\pi/2} cos(x)}= \infty$.

(I just**refuse**to say that "$\displaystyle \frac{1}{0}= \infty$"!) - Jan 10th 2013, 08:19 AMebainesRe: Evaluating arcsec at infinity...?
Go ahead - be pedantic! It keeps us honest.