<s>[INST] Trigonometric Identity question 
$x=2+\csc(\theta)$
$y=\dfrac14\tan(\theta)$
Eliminate $\theta$.

I tried extensively doing $x^2$ and $y^2$ and tried to equate but cannot manage to do it. This is the right method, however.
 [/INST] Do it the hard way:
$x = 2 + \csc \theta = 2 + \frac 1 {\sin \theta}$
$x -2 = \frac 1{\sin \theta}$.  (We can conclude that $\sin \theta \ne 0$ and that $x-2 \ge 1$ or $x - 2 \le -1$ and in any event, $x-2 \ne 0$.)
$\sin \theta =\frac 1{x-2}$
$y = \frac 14{\tan \theta} =  \frac 14 \frac {\sin \theta}{\cos \theta}=\frac 14 \frac {\sin \theta}{\pm\sqrt {1-\sin^2 \theta}}$ (We can conclude $\sin \theta \ne \pm 1$).
$=\frac 14 \frac 1{(x-2)\sqrt{1 - \frac 1{(x-2)^2}}}  $ 
$= \frac 14 \frac 1{\pm\sqrt{(x-2)^2 -1}}$
$16y^2 = \frac 1{x^2 -4x +3}$
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