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.
 A: $$\begin{array}{rcl}
x-2 &=& \csc\theta \\
4y &=& \tan\theta \\
1+\cot^2\theta &=& \csc^2 \theta \\
1+ \dfrac1{16y^2} &=& (x-2)^2
\end{array}$$
A: *

*Start from $x-2 = \frac{1}{\sin \theta}$ and multiply both sides with $4 y = \tan \theta$ to  get $$ 4 y (x-2) = \frac{1}{\cos \theta}$$

*Square both sides.
$$ 16 y^2 (x-2)^2  = \frac{1}{\cos^2\theta} $$


*Use $\underbrace{1+\tan^2 \theta = \frac{1}{\cos^2\theta}}_{\rm trig.\, identity}$ and $\tan\theta = 4 y$ to get


$$ \boxed{ 16 y^2 (x-2)^2 = (4 y)^2+1 } $$
A: 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}}}  $ 
$= \frac 14 \frac 1{\pm\sqrt{(x-2)^2 -1}}$
$16y^2 = \frac 1{x^2 -4x +3}$
