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