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!


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