Which of these parameterize the parabola $y^2=x$? $(\sin^2t,\sin t)$, $(\cos^2t,\cos t)$, $(\sec^2t,\sec t)$, $(\tan^2t,\tan t)$ In this question 

I tried various different methods to solve it but was ending up with the same answer again and again and unfortunately that answer was not given in the options. 
I did my working as shown below

I somehow believe that my answer is correct because if this is a parametric coordinate it should represent any point on this particular parabola.  Now for eg if we want to represent a point (0,0) we can do that as here $\theta$ would be 90 and hence $cot \theta$ = 0 . This doesn't hold true for the answer given in the answer key which is (D) $tan^2(\theta)$ , $tan \theta$ which becomes not defined at $\theta =90$ . 
But I still could be wrong .
It would be greatly appreciated if somebody guides me through this problem.
 A: The problem (I believe) is supposed to be approached like this: we know that a parametric representation for this parabola is obviously $(y^2,y)$ where $y\in{\Bbb R}$. Hence a possible parametrisation (with the change of the parameter $y=f(\theta)$) could be $(f^2(\theta),f(\theta))$ provided that the range of $f$ is $\Bbb R$.
Now look at 4 options and chose that one where the second coordinate ranges over the whole axis. For example, (A) certainly does not work as $y=\sin\theta\in[-1,1]$, which is not $\Bbb R$, so with (A) we get only a part of the parabola, not the whole one, etc
A: As A.Γ said, the solution is obviously in the form $(f(t)^2, f(t))$. 
Let us see if option $A$ works. For the function to be parameterised, then a $t$ must exist for any $k$ in $\sin t = k$. This is to say that if a horizontal line $k$ is drawn, then there must be an intersection between $\sin x$ and $k$ – we are looking for the range of the function.
Then the ranges of the following functions are:
$$\sin x \Rightarrow y \in [-1, 1]$$
$$\cos x \Rightarrow y \in [-1, 1]$$
$$\sec x \Rightarrow y \in [-\infty, -1] \cup [1, \infty]$$
$$\tan x \Rightarrow y \in [-\infty, \infty]$$
Therefore, only $(\tan^2 t, \tan t)$ can parametrise all of $y =x^2$.
