Dear Uncle Colin

These two trig questions are getting me frustrated! What do you recommend?

1. Prove $\frac{\tan(2x) + \cot(x)}{\tan(2x) - \tan(x)} \equiv \cot^2(x)$
2. Prove $\frac{1 + \sin(2x)}{1+\cos(2x)} = \frac{1}{2}\left(1+\tan(x)\right)^2$

- I Don’t Like Equations

Hi, IDLE, and thanks for your message!

The great temptation here is to send you a curt message back saying “What have you tried? What’s holding you up?” - there’s a fine line between “can you help with this?” and “will you do my homework for me?”

Fortunately, the message didn’t reach the top of my inbox in a reasonable timeframe for homework help, so I shall treat it as a request for help with trig generally.

### Question 1

I think the overriding strategy for Nasty Trig is “find ways to make things simpler.”

With the first question, I would look at the left-hand fraction and think about what’s ugly. There are several things that jump out:

• it’s a fraction
• it has a mixture of $2x$ and $x$ in it
• it has a $\cot(x)$ in it.

Let’s deal with the last of those things first. Multiply top and bottom by $\tan(x)$ to get:

$LHS = \frac{\tan(2x)\tan(x) + 1 }{\tan(2x)\tan(x) - \tan^2(x)}$

I can’t say that’s a whole lot nicer. In any case, let’s apply an identity to the $\tan(2x)$, which we know is $\frac{2\tan(x)}{1 - \tan^2(x)}$. In fact, at this point, I will introduce the variable $t$ to be $\tan(x)$.

This gives $LHS = \frac{ \frac{2t^2}{1-t^2} + 1}{\frac{2t^2}{1-t^2} - t^2}$

Ugh. Stacked fractions. We’d better sort those out: multiply top and bottom by $1-t^2$:

$LHS = \frac{ 2t^2 + (1 - t^2) }{2t^2 - t^2(1-t^2)}$

$\dots = \frac{ t^2 + 1}{t^2 + t^4}$

Aha! Now it’s simplifying! If we take the factor of $t^2$ out of the bottom:

$\dots = \frac{t^2 + 1}{t^2(1 + t^2)} = \frac{1}{t^2}$

Which, using our definition, is $\cot^2(x)$ as required.

### Question 2

Starting from $\frac{1 + \sin(2x)}{1+\cos(2x)} = \frac{1}{2}\left(1+\tan(x)\right)^2$, I would again list the ugliness:

• fraction
• mixture of $x$ and $2x$
• mixture of functions

My first step would be to use identities on the left-hand side to see if I can make them any nicer.

$LHS = \frac{1 + 2\sin(x)\cos(x)}{1 + (2\cos^2(x) - 1)}$

Why have I picked that identity for $\cos(2x)$? Because it’ll cancel out with the 1 at the start of the denominator and hopefully make our lives easier.

$\dots = \frac{ 1 + 2\sin(x)\cos(x)}{2\cos^2(x)}$

That’s a start, at least, although we still have a mixture of functions (and no $tan$s, which is what we’re looking for.) How about we divide the $\cos^2(x)$ in?

$\dots = \frac{1}{2} \left( \sec^2(x) + 2\tan(x) \right)$

Oho! A $\sec^2(x)$! We know what to do with those! They’re the same as $1 + \tan^2(x)$ - and again, I’m going to introduce $t$ for $\tan(x)$.

$\dots = \frac{1}{2}\left( 1 + t^2 + 2t \right)$

$\dots = \frac{1}{2}\left( 1 + t\right)^2$, as required.

I hope that helps with your understanding!

- Uncle Colin