Cartesian equation for $x=\sin t \cos t$ and $y=\tan t$

Question

I have already partially asked this question here on this site (Getting the Cartesian equation of parametric equations). Sketchy answers were produced. The answers that i got were

$$ \arctan y= \dfrac{\arcsin(2x)} 2 $$

$$x=y(1-y^2)$$

and

$$ x^2=\dfrac 1 {y^2+1} \left( 1-\left(\dfrac 1 {y^2+1}\right)\right) $$

The graph of the first one of these three cartesian equations was just a portion of the graph that the parametric equation produces while the other two are completely off. What then, is the cartesian equation of this parametric equation (even though some say it might be messy)?

P.S. I could use some help in formatting the equations. Thanks.

Answer 1

hint

With

$$2x=2\sin (t)\cos (t)=\sin (2t) $$

and

$$\sin (2t)=\frac {2\tan (t)}{1+\tan^2 (t)} $$

we get $$xy^2-y+x=0$$

Try to solve for $y $ and find that

$$ y=\frac{1\pm\sqrt {1-4x^2}}{2x} $$

for $x\in [-1/2,1/2] \backslash \{0\}$.

and $x=0\implies y=0$.

Answer 2

With $x=\sin t\,\cos t$ and $y=\tan t$ \begin{cases} xy=\sin t \,\cos t\,\tan t=\sin^2t,\\[2ex] \dfrac{x}{y}=\dfrac{\sin t\cos t}{\tan t}=\cos^2t. \end{cases} so $$\color{blue}{xy+\dfrac{x}{y}=1}$$