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# I was unable to make it to class when this was covered and I'm hoping to see a step-by-step example of how one would write this out.

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I was unable to make it to class when this was covered and I'm hoping to see a step-by-step example of how one would write this out.

Guest Feb 21, 2015

#2
+94105
+13

I always draw a quick pic for these.

let

$$\\\theta=sec^{-1}u\qquad so \\ u=sec\;\theta\\ sec\;\theta = \frac{1}{cos\theta}=\frac{hyp}{adj}=\frac{u}{1}$$

$$cot(sec^{-1}(u))=cot\;\theta = \frac{adj}{opp}=\frac{1}{\sqrt{u^2-1}}$$

NOW u is positive so sec(theta) and cos(theta) are positive. So theta is in the 1st or 4th quadrant.

So cot(theta) can be positive or negative so

$$cot(sec^{-1}(u))=\pm\;\frac{1}{\sqrt{u^2-1}}$$

Melody  Feb 22, 2015
#1
+92674
+8

The arcsec u just means an angle, Θ, whose secant = u.... or just .... u /1

Thus, since the secant ratio is r / x....this means that r = u   and x = 1

And y = √[r^2 - x^2]  = √[u^2 - 1^2] = √[u^2 - 1]

And the cotangent of such an angle is just x / y  =  1 / √[u^2 - 1]    and "rationalizing" the denominator, we have

cot (arcsec u) = cot Θ  = √[u^2 - 1] / [u^2 - 1]

CPhill  Feb 21, 2015
#2
+94105
+13

I always draw a quick pic for these.

let

$$\\\theta=sec^{-1}u\qquad so \\ u=sec\;\theta\\ sec\;\theta = \frac{1}{cos\theta}=\frac{hyp}{adj}=\frac{u}{1}$$

$$cot(sec^{-1}(u))=cot\;\theta = \frac{adj}{opp}=\frac{1}{\sqrt{u^2-1}}$$

NOW u is positive so sec(theta) and cos(theta) are positive. So theta is in the 1st or 4th quadrant.

So cot(theta) can be positive or negative so

$$cot(sec^{-1}(u))=\pm\;\frac{1}{\sqrt{u^2-1}}$$

Melody  Feb 22, 2015
#3
+92674
0

Ah..thanks, Melody....I forgot that this angle could also be in the 4th quad, so I should have added that " -" to my answer...!!!

CPhill  Feb 22, 2015