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
 A: $$\begin{align*}
2\cos 2x\sec x+\sec x\tan x\sin 2x&=2\cos2x\sec x+2\sec x\tan x\sin x\cos x\\
&=2\cos2x\sec x+2\tan x\sin x\\
&=2\left(\frac{\cos2x}{\cos x}+\frac{\sin^2x}{\cos x}\right)\\
&=2\frac{\cos^2x-\sin^2x+\sin^2x}{\cos x}\\
&=\ldots
\end{align*}$$
A: Another way
$$\begin{align}
2\cos x&=\frac{2\cos^2 x}{\cos x}\\
&=\frac{2(\cos^2 x-\sin^2 x+\sin^2 x)}{\cos x}\\
&=\frac{2(\cos2 x+\sin^2 x)}{\cos x}\\
&=2\left(\frac{\cos2 x}{\cos x}+\frac{\sin x\cdot\sin x}{\cos x}\right)\\
&=2(\cos2 x\sec x+\sin x\tan x)\\
&=2\cos2 x\sec x+2\sin x\cdot\frac{\cos x}{\cos x}\tan x\\
&=2\cos2 x\sec x+\sin 2x\sec x\tan x\\
\end{align}$$
A: An alternative way to get to this..... it appears you got the original right hand side by taking the quotient rule of $\frac {\sin (2x)} {\cos x}$.  However,  if you use the double angle identity for sine before taking the derivative, we get 
$\frac {\sin (2x)} {\cos x}=\frac {2 \cos (x) \sin (x)} {\cos x}=2\sin x$
Thus, the derivative of your left hand side is also the derivative of $2 \sin x$, which is $2 \cos x$
As a side note, this is a specific case of a general rule: It's often better to simplify the function BEFORE taking the derivative, as it leads to easier, equivalent answers.
A: Using Double angle formulae, $$\sec x\left(2\cos2x+\tan x\sin2x\right)$$
$$=\sec x\left[2(2\cos^2x-1)+\frac{\sin x}{\cos x}\cdot2\sin x\cos x\right]$$
$$=\sec x\left[4\cos^2x-2+2\sin^2x\right]$$
$$=\sec x\left[4\cos^2x-2(1-\sin^2x)\right]$$
$$=\sec x\left[2\cos^2x\right]=?$$
