We have, L.H.S. = cos $\left(\frac{3\pi}{2}+x\right)$ cos (2π + x) $\left[\cot\left(\frac{3\pi}{2} - x\right) + \cot(2\pi + x)\right]$ = (sin x) (cos x) [tan x + cot x] Since cos $\left(\frac{3\pi}{2}+x\right)$ = sinx , cos (2π + x) = cos x , cot $\left(\frac{3\pi}{2}-x\right)$ = tan x , cot (2π + x) = cot x = cos x sin x $\left[\frac{\sin^2 x + \cos^2 x}{\cos x \sin x}\right]$ = 1 = R.H.S.