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.
 A: 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.
$$
\begin{align*}
1+\tan(x)\tan(2x) &= 1+\frac{\sin(x)}{\cos(x)}\frac{\sin(2x)}{\cos(2x)}\\
&= 1+\frac{\sin(x)2\sin(x)\cos(x)}{\cos(x)\cos(2x)} \quad\text{using the double angle formula for }\sin(2x)\\
&= 1+\frac{2\sin^2(x)}{\cos(2x)}\quad\text{cancelling }\cos(x)\\
&= \frac{\cos(2x)+2\sin^2(x)}{\cos(2x)}\quad\text{getting a common denominator}\\
&= \frac{\cos^2(x)-\sin^2(x)+2\sin^2(x)}{\cos(2x)}\quad\text{using the identity for }\cos(2x)\\
&= \frac{\cos^2(x)+\sin^2(x)}{\cos(2x)}\\
&= \frac{1}{\cos(2x)}\\
&= \sec(2x)
\end{align*}
$$
A: 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.
