# Trig identity $1+\tan x \tan 2x = \sec 2x.$

I need to prove that: $$1+\tan x \tan 2x = \sec 2x.$$

I started this by making sec 1/cos and using the double angle identity for that and it didn't work at all in any way ever. Not sure why I can't do that, but something was wrong.

Anyways I looked at the solutions manual and they magic out $$1 + \tan x \tan 2x = 1+\tan x\left(\frac{2 \tan x}{1-\tan ^2x}\right)$$ which I recognize as the double angle forumla sort of, I just don't understand why I can use that and how they magiced it into this. It is just too difficult to think of an equation in an equation in an equation. I tried working with this and got nowhere near the right answer.

• Since the right hand side does not contain "2x" (but rather, just "x"), you will have to use a double angle identity. I know it seems like magic, but there are some very suggestive hints that you can learn to look for. Sadly, we don't have much time to learn before the test! Jun 24, 2011 at 1:36
• Adam, more people than you realize need to take trig in college! (Even quite a few that studied trig in high school...) Anyway: Nothing to feel ashamed of! Jun 24, 2011 at 1:48

I always like to rewrite things in terms of $\cos$ and $\sin$ when trying to verify identities. I tried to write out the steps in as much detail as possible. \begin{align*} 1+\tan(x)\tan(2x) &= 1+\frac{\sin(x)}{\cos(x)}\frac{\sin(2x)}{\cos(2x)}\\ &= 1+\frac{\sin(x)2\sin(x)\cos(x)}{\cos(x)\cos(2x)} \quad\text{using the double angle formula for }\sin(2x)\\ &= 1+\frac{2\sin^2(x)}{\cos(2x)}\quad\text{cancelling }\cos(x)\\ &= \frac{\cos(2x)+2\sin^2(x)}{\cos(2x)}\quad\text{getting a common denominator}\\ &= \frac{\cos^2(x)-\sin^2(x)+2\sin^2(x)}{\cos(2x)}\quad\text{using the identity for }\cos(2x)\\ &= \frac{\cos^2(x)+\sin^2(x)}{\cos(2x)}\\ &= \frac{1}{\cos(2x)}\\ &= \sec(2x) \end{align*}

• If you cancel $cos(x)$ from top and bottom, you should really also check the case where $cos(x)=0$. Jun 24, 2011 at 15:23

Using the double angle formulas for $\sin$ and $\cos$, we get $$\tan(2x)=\frac{\sin(2x)}{\cos(2x)}=\frac{2\sin(x)\cos(x)}{\cos^2(x)-\sin^2(x)}.$$ Multiplying the top and bottom by $\dfrac{1}{\cos^2(x)}$, we get $$\tan(2x)=\frac{2\sin(x)\cos(x)}{\cos^2(x)-\sin^2(x)}=\frac{\qquad\dfrac{2\sin(x)}{\cos(x)}\qquad}{1-\dfrac{\sin^2(x)}{\cos^2(x)}}=\frac{2\tan(x)}{1-\tan^2(x)}.$$ It is a common trick in mathematics to multiply an expression by $\dfrac{A}{A}$, where $A$ is some clever choice of expresssion. Here we used $A=\dfrac{1}{\cos^2(x)}$. This obviously won't change the value, because $\dfrac{A}{A}=1$ and multiplying by 1 doesn't do anything, but it might collapse some things in what you started with to a form you know how to deal with.

• I am confused tan2x = 2tana divided by 1-tan^2a
• @Adam: Are you familiar with the double-angle formulas for $\sin$ and $\cos$ that I used in the first step? Where in the answer do you get stuck, I will explain more on that step? Jun 24, 2011 at 1:49