<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] Using the formula $$\cos{(a+b)}=\cos{(a)} \cos {(b)}-\sin{(a)} \sin{(b)}$$ we get the following:
$$\cos{ \left ( \frac{3 \pi}{2}+x \right )}=\cos{ \left ( \frac{3 \pi}{2}\right )} \cos{(x)}-\sin{ \left ( \frac{3 \pi}{2} \right )} \sin{(x)}=0 \cdot \cos{(x)}-(-1) \cdot  \sin{(x)}=\sin{(x)}$$
$$\cos{(2 \pi+x)}=\cos{(2 \pi)} \cos{(x)}-\sin{(2 \pi )} \sin{(x)}=\cos{(x)}$$
Use the formula $$\cot{(a+b)}=\frac{\cos{(a)} \cos{(b)}}{\sin{(a)} \cos{(b)}+\cos{(a)} \sin{(b)}}-\frac{\sin{(a)} \sin{(b)}}{\sin{(a)} \cos{(b)}+\cos{(a)} \sin{(b)}} \\ \text{ and }  \\
 \cot{(a-b)}=-\frac{\cos{(a)} \cos{(b)}}{\cos{(a)} \sin{(b)}-\sin{(a)} \cos{(b)}}-\frac{\sin{(a)} \sin{(b)}}{\cos{(a)} \sin{(b)}-\sin{(a)} \cos{(b)}}$$
to calculate $\cot{\left (\frac{3\pi}{2}-x \right )}$ and $\cot{(2 \pi+x)}$.
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