I get a lot of my problems-to-solve from Reddit, since if someone’s posted it there, there are probably thousands of people with the same difficulty. This one isn’t from Reddit, but from @frauenfelder, one of the high-heidyins at BoingBoing.

Out of the 25 homework problems, there was one that she got stuck on. I decided to give it a try and spent two hours on it without solving it.

Here it is. Verify the identity:

$(\sec(x) - \tan(x))^2 \equiv \frac{1 - \sin(x) }{1 + sin(x) }$

(It should really say “for $\sin(x) \ne -1$”, but we’ll roll with it.)

Want to have a go? Be my guest. Spoilers below the line.

Approach 1: start on the left

I’ve done thousands of these over the years, so I have a good sense of what’s going to work. One approach is to start on the left.

$\sec(x) - \tan(x) = \frac{1 - \sin(x)}{\cos(x)}$, which looks promising – we have a $(1-\sin(x))$ on top on the right. But we still have to square it!

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

We have a $\cos^2(x)$ we don’t really want; we can get rid of it with an identity, though: $\cos^2(x) = 1- \sin^2(x)$. So we have our left-hand side as $\frac{(1-\sin(x))^2}{1-\sin^2(x)}$.

Still playing spot the difference, we’ve got a $(1-\sin(x))$ too many on top… but that’s a factor of the bottom, by difference of two squares! So the LHS is $\frac{(1-\sin(x))}{1 + \sin(x)}$, as required.

Approach 2: start on the right

Conjugate trick. I spy a conjugate trick. I don’t like “$1+\sin(x)$” on the bottom. I’m going to multiply everything by $1 - \sin(x)$ to see if it goes away.

The RHS becomes $\frac{(1-\sin(x))^2}{1 - \sin^2(x)}$, or $\frac{(1-\sin(x))^2}{\cos^2(x)}$

That’s $\left( \frac{1 - \sin(x)}{\cos(x)}\right)^2$, and the thing in the brackets is $\sec(x) - \tan(x)$, as required.

I think I prefer the second way, although it relies on an insight. What did you make of it?