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?
 A: $$y=\dfrac{1+\cos2t}{\sin2t}=\cot t$$
$$x=\sec2t+\tan2t=\dfrac{1+\sin2t}{\cos2t}=\dfrac{1+\tan t}{1-\tan t}=\dfrac{y+1}{y-1}$$
Alternatively, $$x\cos\theta-\sin\theta-1=0$$
$$\cos\theta-y\sin\theta+1=0$$
$\implies\cos\theta=?,\sin\theta=?$
A: 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.
A: As is so often the case with trigonometry, the substitution $t=\tan\theta/2$ helps. We have $x=\frac{1+t^2}{1-t^2}+\frac{2t}{1-t^2}=\frac{1+t}{1-t}$ while $y=\frac{1+t^2}{2t}+\frac{1-t^2}{2t}=\frac{1}{t}$, so $x=\frac{1+1/y}{1-1/y}=\frac{y+1}{y-1}$. This function is famously self-inverse, i.e. you can also write $y=\frac{x+1}{x-1}$.
