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 ($-\pi<\theta<\pi$):

Desmos Graphing Calculator - Parametric Equations

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:


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:

Cartesian Equation 1

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:


Plugging that into Desmos returns:

Cartesian Equation 2

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:


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?

  • 2
    $\begingroup$ Just a note for future, use \sin, \cos etc when displaying trig functions, as it comes out better: compare $\sin(x)$ with $sin(x)$ $\endgroup$ – Rhys Hughes May 29 '18 at 11:16
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    $\begingroup$ I think you mean $$x+\frac{1}{x} = 2\sec(θ)\\y+\frac{1}{y} = 2\csc(θ)$$ $\endgroup$ – saulspatz May 29 '18 at 11:27
  • $\begingroup$ @saulspatz Yes, I did. Thank you. $\endgroup$ – Socrates May 29 '18 at 12:30
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    $\begingroup$ when you write $x+\frac{1}{x}=2\sec \theta $, you pick up $x=\sec \theta -\tan \theta $ as well as $x=\sec \theta +\tan \theta $ $\endgroup$ – Lozenges May 29 '18 at 12:49

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

  1. Rewrite everything in terms of $\sin$ and $\cos$.
  2. Try putting things over a common denominator.
  3. 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.

  • $\begingroup$ Thank you very much. That has been very useful. Could you perhaps explain how the solution from $x+\frac{1}{x}$ and the equivalent in $y$ produces spurious solutions? $\endgroup$ – Socrates May 29 '18 at 18:41
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    $\begingroup$ See the comment by @lozenges under your question. It's true that $x = \sec\theta + \tan\theta$ implies $x + \frac{1}{x} = 2\sec\theta$, but this implication is not reversible; in fact $x = \sec\theta - \tan\theta$ also implies $x + \frac{1}{x} = 2\sec\theta$. $\endgroup$ – mweiss May 29 '18 at 18:47

$$y=\dfrac{1+\cos2t}{\sin2t}=\cot t$$

$$x=\sec2t+\tan2t=\dfrac{1+\sin2t}{\cos2t}=\dfrac{1+\tan t}{1-\tan t}=\dfrac{y+1}{y-1}$$

Alternatively, $$x\cos\theta-\sin\theta-1=0$$



  • $\begingroup$ I've followed your first solution through - thank you. However, I don't quite see where you're going with the second. $\endgroup$ – Socrates May 29 '18 at 15:26
  • $\begingroup$ @Socrates, can you eliminate $\theta$? $\endgroup$ – lab bhattacharjee May 29 '18 at 15:41
  • $\begingroup$ That's what I've been trying to do, but I'm not getting anywhere, unfortunately. $\endgroup$ – Socrates May 29 '18 at 16:04
  • $\begingroup$ @Socrates, Use $$\cos^2\theta+\sin^2\theta=1$$ $\endgroup$ – lab bhattacharjee May 30 '18 at 11:51

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 $y=\frac{x+1}{x-1}$.


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