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**+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}$$

Now I can read the positive answer straight off the triangle.

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

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**+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**+13 **

Best Answer

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}$$

Now I can read the positive answer straight off the triangle.

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

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