Written by Colin+ in ask uncle colin.

Dear Uncle Colin,

How do I verify the identity $\frac{\cos(\theta)}{1 - \sin(\theta)} \equiv \tan(\theta) + \sec(\theta)$ for $\cos(\theta) \ne 0$?

- Struggles Expressing Cosines As Nice Tangents

Hi, SECANT, and thanks for your message!

The key questions for just about any trigonometry proof are "what's ugly?" and "how can I clean it up?".

To me, the obvious ugly thing is the $1 - \sin(\theta)$ on the bottom of the fraction. In an *equation*, we'd doubtless multiply both sides through by that and see what dropped out, but this is a *proposed identity* and doing that would involve assuming our answer.

So what can we do instead? I can see two possibilities. One is to use a sort of conjugate trick; the other is to do something *like* multiplying both sides but without actually doing that.

Instead of multiplying both sides by $1-\sin(\theta)$, we can multiply the right-hand side by $\frac{1-\sin(\theta)}{1-\sin(\theta)}$ (so long as $\sin(\theta) \ne 1$ - and we know that's the case, because when $\sin(\theta)=1$, then $\cos(\theta)=0$).

I'd also turn $\tan(\theta)$ and $\sec(\theta)$ into more explicit forms in terms of $\sin(\theta)$ and $\cos(\theta)$, making the right hand side $\br{ \frac{\sin(\theta)}{\cos(\theta)} + \frac{1}{\cos(\theta)} } \frac{1 - \sin(\theta)}{1 - \sin(\theta)}$.

That first bracket can be tidied up to make $\frac{\sin(\theta)+1}{\cos(\theta)}$, so we're left with $\frac{ \br{\sin(\theta) + 1}\br{1 - \sin(\theta)}}{\cos(\theta)\br{1-\sin(\theta)}}$.

Multiplying out the top gives $1 - \sin^2(\theta)$, which is $\cos^2(\theta)$; our right hand side is now $\frac{\cos^2(\theta)}{\cos(\theta)\br{1 - \sin(\theta)}}$. Dividing top and bottom by $\cos(\theta)$ gives the left-hand side we were after! $\blacksquare$

The conjugate trick has a similar flavour to it, but works on the left-hand side. You use this kind of trick when rationalising a denominator -- if the bottom of a fraction is something like $\sqrt{2} + 1$, you multiply top and bottom by $\sqrt{2} -1$, noticing that the result will be the difference of two squares and the square root will magically move to the top.

Here, you have $1 - \sin(\theta)$ on the bottom; multiplying top and bottom by $1 + \sin(\theta)$ will make that nicer. Let's do it:

We get $\frac{\cos(\theta)\br{1+\sin(\theta)}}{\br{1 - \sin(\theta)}\br{1+\sin(\theta)}}$, which is $\frac{\cos(\theta)\br{1+\sin(\theta)}}{1 - \sin^2(\theta)}$. The bottom is identical to $\cos^2(\theta)$; dividing top and bottom by $\cos(\theta)$ leaves us with $\frac{1 + \sin(\theta)}{\cos(\theta)} = \frac{1}{\cos(\theta)} + \frac{\sin(\theta)}{\cos(\theta)}$, or $\sec(\theta) + \tan(\theta)$ as required $\blacksquare$

As you can see, the two methods have a very similar flavour. One requires almost doing something you'd like to do but aren't allowed to; the other requires spotting a trick. Both are valid, and both are useful things to have in your toolbox!

Hope that helps,

- Uncle Colin