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Using trigonometric identities, we get g(x) = tan(x).

For all real numbers x except x=0 and x=1 the function f is defined by f(x/x-1)=1/x . 

Suppose 0<=t<=pi/2. Given that f(sec^(2)t)=(g(theta))^2 for some trigonometric function g, find g.
Enter your answer as one of the six options: sin, cos, tan, sec, csc, cot

$f(\frac{x}{x-1})=\frac{1}{x}\qquad where \;\;\;x\ne0\quad, x\ne1\\ Given\;\; 0<=t <= \frac{\pi}{2}\\ and\\ f(sec^2t)=(g(\theta))^2\qquad \text{Determine the function } g$

[The LaTex less than or equal to command is not working ... that is weird]

I have also found this question confusing:

But this is what I have

$sec^2t=\frac{x}{x-1}\\ xsec^2t-sec^2t=x\\ xsec^2t-x=sec^2t\\ x(sec^2t-1)=sec^2t\\ x=\frac{sec^2t}{sec^2t-1}\\ so\\ f(sec^2t)=\frac{sec^2t-1}{sec^2t}=1-cos^2t=sin^2t\\ so\\g(t)=sin(t)\\ g(\theta)=sin(\theta)$

LaTex

f(\frac{x}{x-1})=\frac{1}{x}\qquad where \;\;\;x\ne0\quad, x\ne1\\   

Given\;\; 0<=t <= \frac{\pi}{2}\\
and\\
f(sec^2t)=(g(\theta))^2\qquad \text{Determine the function  } g

sec^2t=\frac{x}{x-1}\\
xsec^2t-sec^2t=x\\
xsec^2t-x=sec^2t\\
x(sec^2t-1)=sec^2t\\
x=\frac{sec^2t}{sec^2t-1}\\
so\\
f(sec^2t)=\frac{sec^2t-1}{sec^2t}=1-cos^2t=sin^2t\\
so\\g(t)=sin(t)\\
g(\theta)=sin(\theta)