For all real numbers $x$ except $x=0$ and $x=1$ the function $f(x)$ is defined by $f(x/(x-1))=1/x$. Suppose $0\leq t\leq \pi/2$. What is the value of $f(\sec^2t)$?
$\text{(A) } sin^2\theta\quad \text{(B) } cos^2\theta\quad \text{(C) } tan^2\theta\quad \text{(D) } cot^2\theta\quad \text{(E) } csc^2\theta$
Let $y=\frac{x}{x-1} \Rightarrow xy-y=x \Rightarrow x=\frac{y}{y-1}$
$f(y)=\frac{1}{x}=\frac{y-1}{y}=1-\frac{1}{y}$
$f(sec^2t)=sin^2t$
$\fbox{A}$