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
 A: The $\cos(2\pi+x)$ and $\cot(2\pi+x)$ terms are easy to get rid of: the basic trigonometric functions have period $2\pi$. So $\cos(2\pi +x)=\cos x$, and $\cot(2\pi+x)=\cot x$.
As to $\cos(3\pi/2+x)$, we can use the addition formula. It is $\cos(3\pi/2)\cos x-\sin(3\pi/2)\sin x$. This is simply $\sin x$. That's because $\cos(3\pi/2)=0$ and $\sin(3\pi/2)=-1$.
Your turn.  You need to deal with $\cot(3\pi/2 -x)$. Use the procedure of the preceding paragraph, using the fact that $\cot t=\frac{\cos t}{\sin t}$. 
A: 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 $\cot{(2 \pi+x)}$.
A: 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.
$$\require{cancel}\cos \left( \frac{3\pi}2+x \right)=\cancelto{0}{\cos \frac{3\pi}2}\cos x - \cancelto{1}{\sin \frac{3\pi}2}\sin x=\sin x$$
$$\require{cancel}\cos \left(2\pi+x \right)=\cancelto{1}{\cos 2\pi}\cos x - \cancelto{0}{\sin 2 \pi}\sin x=\cos x$$
$$\require{cancel}
\cot\left(\frac{3\pi}2 - x\right) = \frac{\cos(\frac{3\pi}2 - x)}{\sin(\frac{3\pi}2 - x)} = \frac{\cancelto{0}{\cos\frac{3\pi}2} \cos x + \cancelto{1}{\sin\frac{3\pi}2} \sin x }{\cancelto{1}{\sin\frac{3\pi}2} \cos x - \cancelto{0}{\cos\frac{3\pi}2} \sin x} = \frac{\sin x}{\cos x} = \tan x$$
$$\require{cancel}
\cot\left(2\pi + x\right) = \frac{\cos(2\pi + x)}{\sin(2\pi + x)} = \frac{\cancelto{1}{\cos 2\pi} \cos x - \cancelto{0}{\sin 2\pi} \sin x }{\cancelto{0}{\sin 2\pi} \cos x + \cancelto{1}{\cos 2\pi} \sin x} = \frac{\cos x}{\sin x} = \cot x$$
