Trigonometric formula simplifies to $\sin x\cos x[\tan x+\cot x]$ Again, I have a little trouble figuring out how we got from the first step to the next one. It would be really appreciated if someone could help me out. 
$$
\begin{split}LHS &= \cos\left(\frac{3\pi}{2}+x\right)\cos(2\pi +x)\left[\cot\left(\frac{3\pi}{2}-x\right) +\cot(2\pi+x) \right] \\ &= \sin x\cos x[\tan x+\cot x]\end{split}
$$
I've tried the following: $\cot(3\pi/2)$ is $0$, so it would leave out $-\cot x$, that's where I lose the thread. $\cos(3\pi/2)$ is 0, 
so that leaves out $\cos x$ and $\cos2\pi$ is one and that's where I lose the thread.
Note: that this is not an assignment question, these are all solved examples, only for my practice. Source image

Replace the terms on the left side of your left side of the identity with these, and you will end up with the right hand side of your identity.
$$\require{cancel}\cos \left( \frac{3\pi}2+x \right)=\cancelto{0}{\cos \frac{3\pi}2}\cos x - \cancelto{1}{\sin \frac{3\pi}2}\sin x=\sin x$$
$$\require{cancel}\cos \left(2\pi+x \right)=\cancelto{1}{\cos 2\pi}\cos x - \cancelto{0}{\sin 2 \pi}\sin x=\cos x$$
$$\require{cancel}
\cot\left(\frac{3\pi}2 - x\right) = \frac{\cos(\frac{3\pi}2 - x)}{\sin(\frac{3\pi}2 - x)} = \frac{\cancelto{0}{\cos\frac{3\pi}2} \cos x + \cancelto{1}{\sin\frac{3\pi}2} \sin x }{\cancelto{1}{\sin\frac{3\pi}2} \cos x - \cancelto{0}{\cos\frac{3\pi}2} \sin x} = \frac{\sin x}{\cos x} = \tan x$$
$$\require{cancel}
\cot\left(2\pi + x\right) = \frac{\cos(2\pi + x)}{\sin(2\pi + x)} = \frac{\cancelto{1}{\cos 2\pi} \cos x - \cancelto{0}{\sin 2\pi} \sin x }{\cancelto{0}{\sin 2\pi} \cos x + \cancelto{1}{\cos 2\pi} \sin x} = \frac{\cos x}{\sin x} = \cot x$$
