Converting parametric $x = \sec \theta + \tan \theta$, $y = \csc\theta + \cot\theta$ to Cartesian form This question comes from Solomon C4 Paper K, Question 7b.

Consider the parametric equations:
$$\begin{align}
x &= \sec(\theta) + \tan(\theta) \\
y &= \csc(\theta) + \cot(\theta)
\end{align}$$
I would like to express them in Cartesian form. To check my answer, I have used the Desmos Graphing Calculator to draw the graph of the parametric equations ($-\pi<\theta<\pi$):

Firstly I find:
$$\begin{align}
x+\frac{1}{x} &= 2\sec(\theta) \\[4pt]
y+\frac{1}{y} &= 2\csc(\theta)
\end{align}$$
Dividing the first equation by the second:
$$\frac{x+\dfrac{1}{x}}{y+\dfrac{1}{y}}=\tan(\theta)$$
Using both the original parametric equation for $x$ and the one found allows me to simplify to:
$$y+\frac{1}{y} = \frac{2\left(x+\dfrac{1}{x}\right)}{x-\dfrac{1}{x}}$$
At that point, I decide to check with Desmos again:

This graph seems to include the one found from the parametric equations, but with an extra negative reciprocal curve.
At this point, I check the mark scheme for the question, and find that it simplifies to:
$$\begin{align}
\cos(\theta)&=\frac{2x}{x^2+1} \\[4pt]
\sin(\theta)&=\frac{2y}{y^2+1}
\end{align}$$
It then uses the identity: $\sin^2A+\cos^2A=1$ to form the following Cartesian equation:
$$\frac{4y^2}{(y^2+1)^2}+\frac{4x^2}{(x^2+1)^2}=1$$
Plugging that into Desmos returns:

This graph once again includes the original one, but with even more extra reciprocal curves. I think this is because of the squares in the identity used.
I then decide to muck around with various reciprocal equations until I find one that matches the first graph. I find:
$$y=\frac{2}{x-1}+1$$
At this point, I am very confused. How (if it is possible) can I get from my parametric equations to that final Cartesian equation? Where (if I am going wrong) am I (and the mark scheme) going wrong with the first two attempts? Do the first two Cartesian equations have solutions that are not solutions for the parametric equations?

When in doubt, here are some heuristics that can sometimes help:


*

*Rewrite everything in terms of $\sin$ and $\cos$.

*Try putting things over a common denominator.

*Try computing $x+y$, $x^2$, $y^2$ and $xy$ and see if you can find relationships between them.


In this case, we have
$$x = \frac{1}{\cos\theta} + \frac{\sin \theta}{\cos \theta} = \frac{1 + \sin\theta}{\cos \theta}$$
$$y = \frac{1}{\sin\theta} + \frac{\cos \theta}{\sin \theta} = \frac{1 + \cos\theta}{\sin \theta}$$
Finding a common denominator and adding these we have
$$x + y = \frac{\sin\theta + \sin^2\theta + \cos\theta + \cos^2\theta}{\sin\theta\cos\theta} = \frac{1+ \sin\theta +  \cos\theta}{\sin\theta\cos\theta}$$
while multiplication gives
$$xy = \frac{(1+\sin\theta)(1+\cos\theta)}{\sin\theta \cos\theta} = \frac{1+\sin\theta+\cos\theta+\sin\theta \cos\theta}{\sin\theta \cos\theta}$$
Now step back and look at these two results:  hopefully you notice that they are almost identical, except for one extra term in the numerator of the second expression.  In fact we have
$$xy - (x+y) = \frac{\sin\theta \cos\theta}{\sin\theta \cos\theta} = 1$$
so any point on the parametrized curve satisfies the equation
$$xy - x - y = 1$$
Now solve this for $y$ using elementary algebraic methods, and you arrive at the form $$y = \frac{x+1}{x-1}$$
which is equivalent to the equation you found by "mucking around".

As far as why the marking scheme leads to an equation that has "extraneous" solutions: this happens because part of their solution involved squaring both equations.  To see a simpler example of this, suppose the equations were
$$ x = t + 5,   y = t - 5. $$
The most straightforward way to combine these into one equation is to write $y = x - 10$.  However, suppose instead you square both sides, getting:
$$x^2 = t^2 + 10t + 25,   y^2 = t^2 - 10t + 25$$
Then one can observe that $x^2 - y^2 = 20t$.  But also,
$$ x + y = 2t$$
so we can write $$x^2 - y^2 = 10(x + y)$$
This leads to the equations
$$x^2 - 10x - 10y - y^2 = 0$$
which factors into
$$(x - y - 10)(x + y) = 0$$
leading to two separate solutions:
$$y = x - 10 \textrm{  or  } y = -x$$
So the graph of $x^2 - 10x - 10y - y^2 = 0$ consists of two lines, one of which is the one we actually want; the other one is a spurious solution introduced by the act of squaring the original parametric equations.
