trig

\(\csc(\arccos u ) \\~\\ =\frac{1}{\sin(\arccos u)} \\~\\ =\sqrt{ (\frac{1}{\sin(\arccos u)})^2 } \\~\\ =\sqrt{ \frac{1}{\sin^2(\arccos u)} } \\~\\\)

And... sin²θ = 1 - cos²θ

\(=\sqrt{ \frac{1}{1-\cos^2(\arccos u)} } \\~\\ =\sqrt{ \frac{1}{1-\cos(\arccos u)\cos(\arccos u)} } \\~\\ =\sqrt{ \frac{1}{1-uu} } \\~\\ =\frac{1}{\sqrt{1-u^2}} \\~\\\)

This is how I would do it.

let acos(u) = alpha  

Draw a right angled triangle and let alpha be an acute angle.

since  cos( alpha) = u/1

let  the adjacent side by u and the hypotenuse be 1

using pythagoras's theorem you get  

 \(\text{opposite side=}\sqrt{1-u^2}\)

\(cosec(\alpha)=\frac{hyp}{adj}=\frac{1}{\sqrt{1-u^2}}\)

but you do not know which quadrant alpha is in so

\(cosec(\alpha)=\pm\frac{1}{\sqrt{1-u^2}}\)