trig

I think that this is what you mean.

 

\( csc\left[arctan\left(\frac{\sqrt{9-u^2}}{u}\right)\right]\\ \)

 

Draw a right angled triangle.   Mark an acute angle as theta.

\(tan\theta=\frac{\sqrt{9-u^2 } }{u}=\frac{opp}{adj}\\ hyp=\sqrt{9-u^2+u^2}=3\)

 

\(csc\theta = \frac{hyp}{opp}=\frac{3}{\sqrt{9-u^2}}\)

 

But u can be positive or negative     -3

 

so 

 

\( csc\left[arctan\left(\frac{\sqrt{9-u^2}}{u}\right)\right]=\pm\frac{3}{\sqrt{9-u^2}} \)

 

I think that is ok :)