I mean, I don’t mean to imply that I’m a master, although I’m pretty good at A-level trig. I just wanted to talk through my thought process in solving a question, and “masterclass” seemed like a good word.

The question is:

(a) Show that $\frac{1}{\cos(\theta)} +\tan(\theta) \equiv \frac{\cos(\theta)}{1-\sin(\theta)},\quad \theta \ne (2n+1)90^\circ, \quad n \in \mathbb Z$

Given that $\cos(2x) \ne 0$

(b) Solve for $0 < x < 90^\circ$:

$\frac{1}{\cos(2x)} + \tan(2x) = 3 \cos(2x)$

It’s three marks for (a) and five for (b), which I don’t really care about, but tends to give an idea of how much work is expected.

OK! Let’s start with the show that bit.

My first thought is, $\tan(\theta)$ is the same as $\frac{\sin(\theta)}{\cos(\theta)}$, which makes the left-hand side quite simple – but an annoyingly different form to the right-hand side. I’ll work it out: the left-hand side is $\frac {1}{\cos(\theta)} + \frac{\sin(\theta)}{\cos(\theta)}$, or $\frac{1 + \sin(\theta)}{\cos(\theta)}$.

That’s not (immediately) the same as $\frac{\cos(\theta)}{1-\sin(\theta)}$, though. I could check, if I wanted – put in a random value of $\theta$ on both sides and see if I get the same. If I’ve made an error, this will most likely tell me about it. It doesn’t prove anything, though.

My spidey senses tingle at seeing both a $1-\sin(\theta)$ and a $1 + \sin(\theta)$ – multiplying those would give me a $\cos^2(\theta)$, and a little bit of mental cross-multiplication convinces me that that’s exactly what we’d get. Unfo rtunately, that’s not a permitted move in a show that question.

What is permitted is to multiply top and bottom by the same, non-zero thing: so I can do something like $\frac{1+\sin(\theta)}{\cos{\theta}}\cdot\frac{1-\sin(\theta)}{1-\sin(\theta)}$ – so long as $\sin(\theta)\ne 1$. Fortunately, that ‘s what the condition after the equivalence means, so it’s legit. Multiplying it out gives $\frac{(1+\sin(\theta))(1-\sin(\theta)}{\cos(\theta)(1-\sin(\theta))}$, or $\frac{1 - \sin^2(\theta)}{\cos(\theta)(1-\sin(\theta)}$.

We’re nearly there! If we replace the top with $\cos^2(\theta)$ and cancel a $\cos(\theta)$– which we can do, since the condition also implies that $\cos(\theta) \ne 0$ – we get $\frac{\cos(\theta)}{1-\sin(\theta)$. Boom!

The second part does something a lot of questions like this do: reuse the result from the first part in the second. (It’s worth noting that you can use the result even if you didn’t get part (a) – you don’t have to miss out on these marks just because you didn’t get the first bit right.) It’s the same sort of shape – you just need to play spot the difference and notice that $\theta$ has become $2x$.

I’d prefer to use $\theta$ – and since $0 < x < 90^\circ$, we can say $0 < \theta < 180^\circ$ (since it’s the same as $2x$.

Now rewrite it as $\frac{1}{\cos(\theta) + \tan(\theta) = 3 \cos(\theta)$ and replace the LHS with the result from earlier: $\frac{\cos(\theta)}{1-\sin(\theta)} = 3 \cos(\theta)$.

We’re told that $\cos(\theta) \ne 0$, so we can divide it out: $\frac{1}{1-\sin(\theta)} = 3$, or $1 - \sin(\theta) = \frac{1}{3}$. Better yet, $\sin(\theta) = \frac{2}{3}$, and the calculating machine gives an answer of $41.8^\circ$ for $\theta$. 1

It would be easy – and wrong – to write down 41.8 degrees and think “job’s a good-un”.

It would be easy – and incomplete – to notice that we made $\theta$ up and we need to halve it to get $x$, and write down $20.9^\circ$.

The correct thing to do is to draw a small graph of $y=\sin(\theta)$ and notice that there’s a second solution at $\theta = (180-41.8)^\circ$. That works out to $x = (90 - 20.9)^\circ$, or $69.1^\circ$.

There are two solutions: 20.9 and 69.1 degrees, and it’s good practice to put them (or better, the exact values) back into the equation to check.

Would you have done it differently? Was there anything I missed? Do let me know!

  1. I pretend it was the calculating machine. It’s actually a value I have memorised in case the Ninja is around. ↩