Verify $\sec(2x) = (\sec^2(x))/(2-\sec^2(x))$ Again, can someone give me just a hint as to how to start this identity verification? I've tried subbing in the all 3 cos double angle identities to the left side, but it keeps ending up back at the start.
$$\sec(2x) = \frac{\sec^2(x)}{2-\sec^2(x)}$$
 A: You are then familiar with the identity
$$\cos(2x)=2\cos^2(x)-1.$$
From this we obtain
$$\frac{1}{\cos(2x)}=\frac{1}{2\cos^2(x)-1}.$$
By the definition of $\sec$, we have then
$$\sec(2x)=\frac{1}{2\cos^2(x)-1}.$$
Stop reading here and do the rest on your own.  If you experience trouble, continue reading.
Now on the right-hand side, divide top and bottom by $\cos^2(x)$.
When we divide $1$ by $\cos^2(x)$, we get $\sec^2(x)$.
When we divide $2\cos^2(x)-1$ by $\cos^2(x)$, we get $2-\sec^2(x)$. That completes the solution. 
Comment: I think of the "three identities" as a single identity, which you can manipulate using $\sin^2x+\cos^2 x$ into various equivalent forms.
But thinking about the three identities that you have seen for $\cos(2x)$, let's see which one should be our first pick to manipulate.  The $2$ in the given right-hand side suggests that it should be one of the identities that involves a $2$ somwhere.  And since your given right-hand side involves only a close relative of $\cos x$, I would choose $\cos(2x)=2\cos^2(x)$ for manipulation.
The manipulation can be done in various ways, but probably should use $\sec u=1/\cos(u)$.
So for example, we can first write $2\cos^2 x -1$ as $\dfrac{2}{\sec^2 x}-1$, and rewrite that as $\dfrac{2-\sec^2(x)}{\sec^2(x)}$.
A: First, $\sec 2x=\frac{1}{\cos 2x}$. Write $1$ as $\cos^2  x +\sin ^2 x$ and $\cos 2x$ as $\cos^2 x-\sin^2x$. Then $$\sec 2x=\frac{\cos^2 x+\sin^2x}{\cos^2 x-\sin^2x}.$$ Next, divide top and bottom (of the right hand side) by $\cos^2x$ and use the fact that $1+\tan ^2x =\sec^2x$.
A: Note that $$\sec(2x) = \frac{1}{\cos(2x)}$$ and $$\cos(2x) = \frac{1- \tan^{2}x}{1+ \tan^{2}x}$$
