How do I go from 1-(tan^2)(arctan x/y ) to 1 - (x^2)/(y^2)
Functions tan and arctan are inverse of each other. More precisely, tan is the left inverse of arctan everywhere, i.e., tan(arctan(x)) = x for all x and tan is the right inverse of arctan for $\displaystyle x\in(-\pi/2,\pi/2)$, i.e., arctan(tan(x)) = x on this interval.
I think what is confusing me is, why does the square on the tan^2(arctan....) etc get distributed to the x/y or is it more complicated than it being distributed. Also is arctan the same as 1/tan ? Thanks for the help, I am sorry if I confuse you but I want to know this well.
The notation $\displaystyle \tan^2(x)$ means $\displaystyle (\tan(x))^2$ by definition. So, $\displaystyle \tan^2(\arctan(x/y))=(\tan(\arctan(x/y)))^2=(y/x)^2$.
No, these are two very different things. See the graphs of the two functions here. For a given x, 1/tan(x) = cot(x) is the reciprocal of tan(x) whereas $\displaystyle \tan^{-1}(y)=\arctan(y)$ is the unique $\displaystyle x\in(\pi/2,\pi/2)$ such that tan(x) = y. Note that by analogy with $\displaystyle \tan^2(x)=(\tan(x))^2$, $\displaystyle \tan^{-1}(y)$ should denote $\displaystyle (\tan(y))^{-1}=1/\tan(y)$, but this is not so. Thus, $\displaystyle \tan^n(x)$ denotes $\displaystyle (\tan(x))^n$ for n > 0, but $\displaystyle \tan^{-1}(x)$ usually denotes arctan(x), the inverse function of tan(x). This is somewhat confusing, but there is a tradition of denoting the inverse of a function $\displaystyle f$ by $\displaystyle f^{-1}$.