Show that $2\cos(x)$ is equal to $2\cos(2x)\sec(x)+\sec(x)\tan(x)\sin(2x)$ This is from the derivative of $\dfrac{\sin(2x)}{\cos x}$
I tried to solve it and arrived with factoring the $\sec(x)$ but I still can't get it to $2\cos(x)$. Could you help me out, please? Thanks

Another way
$$\begin{align}
2\cos x&=\frac{2\cos^2 x}{\cos x}\\
&=\frac{2(\cos^2 x-\sin^2 x+\sin^2 x)}{\cos x}\\
&=\frac{2(\cos2 x+\sin^2 x)}{\cos x}\\
&=2\left(\frac{\cos2 x}{\cos x}+\frac{\sin x\cdot\sin x}{\cos x}\right)\\
&=2(\cos2 x\sec x+\sin x\tan x)\\
&=2\cos2 x\sec x+2\sin x\cdot\frac{\cos x}{\cos x}\tan x\\
&=2\cos2 x\sec x+\sin 2x\sec x\tan x\\
\end{align}$$
