<s>[INST] Prove $\frac{2\cos x}{\cos 2x + 1 }= \sec x$ Prove that $\dfrac{2\cos x}{\cos 2x + 1 }= \sec x$. 
So far I have:
$\dfrac{2\cos x}{\cos 2x + 1 }= \dfrac 1 {\cos x}$
Where do I go from here?
 [/INST] It seems to be a widespread practice for students to begin attempted proofs of trigonometric identities by writing things like this:
$$\frac{2\cos x}{\cos 2x + 1 }= \sec x$$
$$\frac{2\cos x}{\cos 2x + 1 }= \frac 1 {\cos x}$$
That's ok in scratchwork, but the finished proof should go like this:
$$
\sec x = \frac 1 {\cos x} = \cdots \cdots \cdots = \frac{2\cos x}{\cos2x + 1}
$$
or like this:
$$
\frac{2\cos x}{\cos2x + 1} = \cdots \cdots \cdots = \frac 1 {\cos x} = \sec x
$$
and of course to finish it you need to figure out what goes where all those dots are.
In other words, you put $\text{“}=\text{''}$ between things that you already know are equal.
Seeing $\cos(2x)$, you should recall the double-angle formula that says $\cos(2x) = \cos^2 x - \sin^2 x$.  Then you have
\begin{align}
\frac{2\cos x}{\cos(2x)+1} & = \frac{2\cos x}{\cos^2 x - \sin^2 x + 1} \\[10pt]
& = \frac{2\cos x}{\cos^2 x + \cos^2 x} & & \text{since }-\sin^2x+1 = \cos^2 x \\[10pt]
& = \frac{2\cos x}{2\cos^2 x} \\[10pt]
& = \frac 1 {\cos x} \\[10pt]
& = \sec x.
\end{align}
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