Prove that $\cos \left(\frac{3 \pi}{2}+x\right) \cos (2 \pi+x)\left[\cot \left(\frac{3 \pi}{2}-x\right)+\cot (2 \pi+x)\right]=1$.

(N/A) $L.H.S. = \cos \left(\frac{3 \pi}{2}+x\right) \cos (2 \pi+x) \left[\cot \left(\frac{3 \pi}{2}-x\right)+\cot (2 \pi+x)\right]$
Using allied angle formulas:
$\cos \left(\frac{3 \pi}{2}+x\right) = \sin x$
$\cos (2 \pi+x) = \cos x$
$\cot \left(\frac{3 \pi}{2}-x\right) = \tan x$
$\cot (2 \pi+x) = \cot x$
Substituting these values:
$= \sin x \cos x [\tan x + \cot x]$
$= \sin x \cos x \left(\frac{\sin x}{\cos x} + \frac{\cos x}{\sin x}\right)$
$= \sin x \cos x \left[\frac{\sin^2 x + \cos^2 x}{\sin x \cos x}\right]$
$= \sin^2 x + \cos^2 x = 1 = R.H.S.$