We start with the left-hand side (LHS) of the equation:

Simplifying $\cos(3\pi/2 + x)$: Using the cosine addition formula, $\cos(A+B) = \cos A \cos B - \sin A \sin B$, we have

$$\cos\left(\frac{3\pi}{2} + x\right) = \cos\left(\frac{3\pi}{2}\right) \cos x - \sin\left(\frac{3\pi}{2}\right) \sin x$$

Since $\cos(3\pi/2) = 0$ and $\sin(3\pi/2) = -1$, this becomes

$$0 \cdot \cos x - (-1) \cdot \sin x = \sin x$$

Simplifying $\cos(2\pi + x)$: Similarly,

$$\cos(2\pi + x) = \cos(2\pi) \cos x - \sin(2\pi) \sin x$$

With $\cos(2\pi) = 1$ and $\sin(2\pi) = 0$, we get

$$1 \cdot \cos x - 0 \cdot \sin x = \cos x$$

Simplifying $\cot(3\pi/2 - x)$: Write $\cot(3\pi/2 - x) = \cos(3\pi/2 - x)/\sin(3\pi/2 - x)$. Using the cosine and sine difference formulas: $\cos(A-B) = \cos A \cos B + \sin A \sin B$, and $\sin(A-B) = \sin A \cos B - \cos A \sin B$. Thus,

$$\cos\left(\frac{3\pi}{2} - x\right) = \cos\left(\frac{3\pi}{2}\right) \cos x + \sin\left(\frac{3\pi}{2}\right) \sin x = 0 \cdot \cos x + (-1) \cdot \sin x = -\sin x$$

And

$$\sin\left(\frac{3\pi}{2} - x\right) = \sin\left(\frac{3\pi}{2}\right) \cos x - \cos\left(\frac{3\pi}{2}\right) \sin x = (-1) \cdot \cos x - 0 \cdot \sin x = -\cos x$$

Hence,

$$\cot\left(\frac{3\pi}{2} - x\right) = \frac{-\sin x}{-\cos x} = \tan x$$

Simplifying $\cot(2\pi + x)$: Since the cotangent function has period $\pi$, $\cot(2\pi + x) = \cot x$. Alternatively, using the definitions:

$$\cot(2\pi + x) = \frac{\cos(2\pi + x)}{\sin(2\pi + x)} = \frac{\cos x}{\sin x} = \cot x$$

Combining the terms: Now the LHS becomes $\sin x \cdot \cos x \cdot [\tan x + \cot x]$. Express $\tan x = \sin x/\cos x$ and $\cot x = \cos x/\sin x$. Then

$$\tan x + \cot x = \frac{\sin^2 x + \cos^2 x}{\sin x \cos x}$$

By the Pythagorean identity, $\sin^2 x + \cos^2 x = 1$, so

$$\tan x + \cot x = \frac{1}{\sin x \cos x}$$

Therefore,

$$\text{LHS} = \sin x \cos x \cdot \frac{1}{\sin x \cos x} = 1$$

which equals the right-hand side (RHS).

Hence, the given identity is proved. ∴ LHS = RHS = 1