A puzzle from BoingBoing

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.

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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!

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

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.

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