How is $\frac{dt}{dx}$ just the reciprocal of $\frac{dx}{dt}$ ?? Came across this problem.  Derivative of a parametric.
$$x=2\cos(t)$$
$$y=2\sin(t)$$
.....Hence,
$$\frac{dx}{dt}=-2\sin(t)$$
$$\frac{dy}{dt}=2\cos(t)$$
As we know, chain rule states:   $$\frac{dy}{dx} = \frac{dy}{dt} \cdot \frac{dt}{dx}$$
Apparently, this results in  $$\frac{dy}{dx} = 2\cos(t)\frac{1}{-2\sin(t)}$$
That last part seems weird.  How do you get $\frac{dt}{dx} = \frac{1}{-2\sin(t)}$ ?  You just take the reciprocal of $\frac{dx}{dt}$ ??   Seems like a bizarre coincidence.    Inverse is not the same as reciprocal.
If I wanted to get $\frac{dt}{dx}$ explicitly, start with the original:
$$x=2\cos(t)$$
Then put in terms of t, and you'd have:
$$t = \cos^{-1}(\frac{x}{2})$$
Then take $$\frac{dt}{dx} = - \frac{1}{\sqrt{1-\bigl(\frac{x}{2}\bigr)^2}} \cdot \frac{1}{2} = -\frac{1}{\sqrt{4-x^2}}$$
The point is this doesn't look anything like what we used for $\frac{dt}{dx}$ which was $\frac{1}{-2\sin(t)}$.

Your last step is wrong. It should be
$$\frac{dt}{dx}=-\frac12\cdot\frac{1}{\sqrt{1-\frac{x^2}{4}}} $$ $$\frac{dt}{dx}=-\frac12\cdot\frac{1}{\sqrt{1-\cos^2 t}} $$
A much better approach would be to differentiate    $\space x=2\cos t $ with respect to $x $
$$1=-2\sin t \frac{dt}{dx} $$
