<s>[INST] 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
 [/INST] 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.
$$\require{cancel}\cos \left( \frac{3\pi}2+x \right)=\cancelto{0}{\cos \frac{3\pi}2}\cos x - \cancelto{1}{\sin \frac{3\pi}2}\sin x=\sin x$$
$$\require{cancel}\cos \left(2\pi+x \right)=\cancelto{1}{\cos 2\pi}\cos x - \cancelto{0}{\sin 2 \pi}\sin x=\cos x$$
$$\require{cancel}
\cot\left(\frac{3\pi}2 - x\right) = \frac{\cos(\frac{3\pi}2 - x)}{\sin(\frac{3\pi}2 - x)} = \frac{\cancelto{0}{\cos\frac{3\pi}2} \cos x + \cancelto{1}{\sin\frac{3\pi}2} \sin x }{\cancelto{1}{\sin\frac{3\pi}2} \cos x - \cancelto{0}{\cos\frac{3\pi}2} \sin x} = \frac{\sin x}{\cos x} = \tan x$$
$$\require{cancel}
\cot\left(2\pi + x\right) = \frac{\cos(2\pi + x)}{\sin(2\pi + x)} = \frac{\cancelto{1}{\cos 2\pi} \cos x - \cancelto{0}{\sin 2\pi} \sin x }{\cancelto{0}{\sin 2\pi} \cos x + \cancelto{1}{\cos 2\pi} \sin x} = \frac{\cos x}{\sin x} = \cot x$$
</s>