Find an expression for :
csc^2 (tan^-1 (3/x))
in terms of x
I think this is what the problem looks like?
$\displaystyle csc^2(Arctan(\frac{3}{x}))$
Arctan is just one way of writing the inverse tangent function.
to solve this, let $\displaystyle \theta=Arctan(\frac{3}{x})$.
Therefore, $\displaystyle tan(\theta)=\frac{3}{x}$
and
$\displaystyle csc^2(Arctan(\frac{3}{x}))=csc^2(\theta)=1+cot^2(\ theta)=1+\frac{1}{tan^2(\theta)}$
So now you can write this in terms of
$\displaystyle tan(\theta)=\frac{3}{x}$
$\displaystyle csc^2(Arctan(\frac{3}{x}))=1+\frac{1}{\frac{9}{x^2 }}=1+\frac{x^2}{9}$
Note, I edited my original post because I messed up on the final line. So if you already looked at it, look again. It's fixed now. That should be the expression you're looking for.
- BTW, this kind of question probably should have been posted in the trignometry section, since these are inverse trig functions.