Eliminate the parameter given $x = \tan^{2}\theta$ and $y=\sec\theta$

Question

$x = \tan^{2} (\theta)$ and $y = \sec (\theta)$

knowing that $\tan^{2} (\theta) = (\tan (\theta))^2 = \dfrac{\sin^{2}\theta}{\cos^{2}\theta}$

and that $\sec(\theta) = \dfrac{1}{\cos(\theta)}$ $\to$ $y=\dfrac{1}{\cos(\theta)}$

For $y$ we can get an even more simplified expression that will be useful for substituting into $x$ by multiplying both sides by $\cos(\theta)$ and then dividing both sides by $y$ to obtain: $\cos(\theta)=\frac{1}{y}$

With the trig simplified throught identities know we can now use more identities that become obvious and then simplify to a final expression through substitution into the x component:

$x = \dfrac{1-(\cos(\theta))^{2}}{(\cos(\theta))^{2}}$

Now substituting $\cos(\theta)$ for $\frac{1}{y}$ we get:

$x = \dfrac{1-(\frac{1}{y})^2}{(\frac{1}{y})^2}$

which simplifies to:

$x = \dfrac{1-\frac{1}{y^2}}{\frac{1}{y^2}}$

which can be simplified to:

$x = \dfrac{1-(\frac{1}{y^2})}{1} \cdot \dfrac{y^2}{1} \to y^2 - 1$

So removing the parameter we get:

$x= y^2 - 1$

I feel very confident that this answer is correct but I would like verification.

Answer 1

Recall the circular identity $$\sin^2 \theta + \cos^2 \theta = 1.$$ Dividing both sides by $\cos^2 \theta$ gives $$\tan^2 \theta + 1 = \sec^2 \theta.$$ Consequently, $$x + 1 = y^2.$$

Answer 2

You can save some passages: $$ x=\frac{\sin^2\theta}{\cos^2\theta}=\frac{1-\cos^2\theta}{\cos^2\theta}= \frac{1}{\cos^2\theta}-\frac{\cos^2\theta}{\cos^2\theta}=y^2-1 $$