# Ask Uncle Colin: A Trigonometric Identity

Dear Uncle Colin,

I got stuck trying to prove that $(\sec(x) - \tan(x))^2 \equiv \frac{1-\sin(x)}{1+\sin(x)}$. Can you help?

- Bringing Over Is No Good! Bringing Over Is No Good!

Hi, BOINGBOING, and thank you for your message!

Here’s how I’d tackle it, starting from the left hand side:

- $\sec(x) - \tan(x) = \frac{1 - \sin(x)}{\cos(x)}$
- $(\sec(x) - \tan(x))^2 = \frac{ (1 -\sin(x))^2}{\cos^2(x)}$
- $\dots = \frac{(1-\sin(x))^2}{1-\sin^2(x)}$
- $\dots = \frac{(1-\sin(x))^2}{(1-\sin(x))(1+\sin(x))}$ by difference of two squares
- $\dots = \frac{1-\sin(x)}{1+\sin(x)} \blacksquare$

I think the nugget to this one is converting the $\cos^2(x)$ into something involving sine, and then using the difference of two squares on that.

### But what if you don’t see the nugget?

It’s drilled into you, from the moment you start proofs, that you’re not to treat the thing you’re trying to prove like an equation. You’re not to move things around willy-nilly.

*But this only applies to your final answer!* When you’re exploring and trying to find your way through, pretty much anything goes. So, if you messed around with it like so:

- $(\sec(x) - \tan(x))^2 = \frac{1-\sin(x)}{1+\sin(x)}$
- $\left( \frac{1 - \sin(x)}{\cos(x)} \right)^2 = \frac{1-\sin(x)}{1+\sin(x)}$
- $\frac{1-\sin(x)}{\cos^2(x)} = \frac{1}{1+\sin(x)}$
- $(1-\sin(x))(1+\sin(x)) = \cos^2(x)$

You would think: aha! I know *that’s* true! And you can then use this mess as a hint for how to proceed. The only ‘illegal’ thing we’ve done here is in the last couple of lines, when we started to cancel. If, instead, we started to factorise on the left, we’d have:

- $(\sec(x) - \tan(x))^2 \dots$
- $\dots = \left( \frac{1 - \sin(x)}{\cos(x)} \right)^2$
- $\dots = \frac{(1-\sin(x))(1-\sin(x))}{\cos^2(x)}$
- $\dots = \frac{(1-\sin(x))(1-\sin(x))}{(1-\sin(x))(1+\sin(x))}$

… which is eerily similar to what we did first time out!

Hope that helps!

- Uncle Colin

* Fixed typo in title, 2020-02-26. Thanks, Adam!