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 , 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.
here. For a given x, 1/tan(x) = cot(x) is the reciprocal of tan(x) whereas is the unique such that tan(x) = y. Note that by analogy with , should denote , but this is not so. Thus, denotes for n > 0, but 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 by .