Dear Uncle Colin,

I have a trig identity I can’t prove! I have to show that $\frac{\cos(x)}{1-\sin(x)} = \tan(x) + \sec(x)$.

Strangely Excited Comment About Non-Euclidean Trigonometry.

Hi, SECANT, and thanks for your message!

This is a slightly sneaky one, but definitely a good one to practice. Let’s do it the normal way, then look at the mess I’ve glossed over on the way.

### The proof

First, working on the right hand side, multiply top and bottom by $(1+\sin(x))$. That makes it $\frac{(\cos(x))(1 + \sin(x))}{(1-\sin(x))(1 + \sin(x))}$.

The bottom of the fraction is $\cos^2(x)$, and a bit of simplifying gives $\frac{1}{\cos(x)} + \frac{\sin(x)}{\cos(x)} = {\sec(x) + \tan(x)}$, as required. $\blacksquare$

### But where did you pull $(1+\sin(x))$ out from?

From my vast experience, dear reader. Perhaps a better question is, how can you pull out a $(1+\sin(x))$ when needed?

One way to figure this out is to do things that are a tiny bit dicey in terms of a Proper Proof, but are really useful hacks in the figuring-out stage.

If we look at the right-hand side instead of the left, we have $\frac{\sin(x)+1}{\cos(x)}$, which needs to be equivalent to $\frac{\cos(x)}{1-\sin(x)}$. We can convince ourselves that that’s true by carefully multiplying each side by the bottom of its fraction. Doing that gives us $1-\sin^2(x) = \cos^2(x)$, which we know to be true.

It also suggests that an important step on the way is to bring a $\sin(x)+1$ onto the left-hand side - which indeed it is. A way to do that, while keeping the left-hand expressions equivalent to each other, is to multiply top and bottom by the same expression - just as we did!

### Careful

The question as you’ve stated it is incomplete, though. It should also specify that the identity only holds so long as $\cos(x) \ne 0$, to avoid any of the functions going to infinity.

Hope that helps!

- Uncle Colin