How do you convert \[r=\tan \theta \times \sec \theta \] into Cartesian form?

Hint: Here we will use the parametric relations \[x=r\cos \theta\] and \[y=r\sin \theta \] to find the value of $\tan \theta $ and substitute it in the given relation. Here, x and y denotes the x – coordinate and y – coordinate in Cartesian form respectively. Now, we will use the conversion $\sec \theta =\dfrac{1}{\cos \theta }$ and substitute the value of \[\cos \theta \] in terms of x and r. finally we will cancel the common terms to get the required equation in Cartesian form.

Complete step by step solution:
We can see that the given relation is in polar form because it relates the radius vector (r) and angle (\[\theta\]). We need to change it into Cartesian form that means we have to obtain the relationship between x and y – coordinate.
If we consider a point (P) with Cartesian coordinates P (x, y) and polar coordinates \[\left( r,\theta \right)\] then the relation between the two-coordinate system is given as: -
\[\Rightarrow x=r\cos \theta\] and \[y=r\sin \theta\]
Considering the first relation we have,
$\Rightarrow \cos \theta =\dfrac{x}{r}................\left( i \right)$
Also, dividing y with x we get,
$\Rightarrow \dfrac{y}{x}=\dfrac{sin\theta }{\cos \theta }$
We know that the ratio of sine and cosine function is the tangent function, so we get,
$\Rightarrow \dfrac{y}{x}=\tan \theta ................\left( ii \right)$
Now, let us come to the given polar equation. So, we have,
\[\Rightarrow r=\tan \theta \times \sec \theta \]
Using equations (ii) we get,
\[\Rightarrow r=\dfrac{y}{x}\times \sec \theta \]
Now, using the relation: $\sec \theta =\dfrac{1}{\cos \theta }$ and equation (i) we get,
\[\begin{align}
  & \Rightarrow r=\dfrac{y}{x}\times \dfrac{1}{\cos \theta } \\
 & \Rightarrow r=\dfrac{y}{x}\times \dfrac{1}{\left( \dfrac{x}{r} \right)} \\
 & \Rightarrow r=\dfrac{y}{x}\times \dfrac{r}{x} \\
 & \Rightarrow r=\dfrac{r\times y}{{{x}^{2}}} \\
\end{align}\]
By cross – multiplication and cancelling the common factor we get,
\[\Rightarrow {{x}^{2}}=y\]
Hence, the above relation represents the given equation in Cartesian form.

Note: If we are asked the type of curve the obtained relation represents then our answer will be an upward parabola. Here you may think that why we have cancelled the common factor r from both the sides. The reason is that the relation between r, x and y is $r=\sqrt{{{x}^{2}}+{{y}^{2}}}$ so if we will take r common and substitute it equal to 0 then we will get the relation \[{{x}^{2}}+{{y}^{2}}=0\] which is only possible when the point lies on the origin. You may consider this particular case because in other cases, the obtained relation will be undefined as we are considering real numbers only.