How to simplify $\cot(\sec^{-1}(e^x))$

Question

I've been trying to simplify $\cot(\sec^{-1}(e^x))$. I thought substitution might the way to go about it so I said:

let $u = \sec^{-1}({e^x})$

I'm therefore trying to find $\cot(u)$

From $u = \sec^{-1}({e^x})$ I can say:

$\sec(u) = e^x$

$\therefore \cos(u) = \dfrac{1}{e^x} $

$\therefore \cot(u) = \dfrac{1}{e^x\sin(x)}$

However, I want my answer just in terms on exponentials and I'm not too sure how to go about it after this point?

Thank you :)

Answer 1

If $\sec^{-1}(e^x)=u,0\le u\le\pi$ based on the definition of Principal value

But, $\displaystyle e^x>0$ for finite real $x$

$\displaystyle\implies 0\le u<\frac\pi2 $

$\implies \cot u\ge0$

Now as $\displaystyle\cot^2u=\frac1{\tan^2u}=\frac1{\sec^2u-1}$

$\displaystyle\implies\cot u=+\frac1{\sqrt{\sec^2u-1}}$