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