# Parametric Curves finding its cartesian equation [closed]

A curve has parametric equations: $x=2\csc(X)$, $y=\cot(X)$. How do I find the cartesian equation of the curve? Thanks in advance.

## closed as off-topic by José Carlos Santos, J. M. is a poor mathematician, Claude Leibovici, user26857, user91500Aug 28 '17 at 9:53

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My first reaction to any problem involving trig functions is to write them in terms of sine and cosine. Here, $x= 2 csc(X)= \frac{2}{sin(X)}$ and $y= \frac{cos(X)}{sin(X)}$. From the first equation $sin(X)= \frac{2}{x}$ so that $cos(X)= \sqrt{1- sin^2(X)}= \sqrt{1- \frac{4}{x^2}}= \frac{\sqrt{x^2- 4}}{x}$
$y= \frac{\frac{\sqrt{x^2- 4}}{x}}{\frac{2}{x}}= \frac{1}{2}\sqrt{x^2- 4}$.
To allow for possible sign changes, multiply both sides by 2 and square: $4y^2= x^2- 4$ which reduces to the hyperbola $x^2- 4y^2= 4$.
squaring your second equation we get $$y^2=\frac{1-\sin^2(t)}{\sin^2(t)}$$ and from $$\sin(t)=\frac{2}{x}$$ we obtain $$y^2=\frac{1-\left(\frac{2}{x}\right)^2}{\left(\frac{2}{x}\right)^2}$$