Step 1: Evaluate each term individually

We solve the expression term by term using the properties of inverse trigonometric functions.

Term 1: tan^-1(-1/√3). Since tan^-1(-x) = -tan^-1(x), we get: $$-\tan^{-1}\left(\frac{1}{\sqrt{3}}\right) = -\frac{\pi}{6}$$

Term 2: cot^-1(1/√3). Since cot(π/3) = 1/√3, we get: $$\cot^{-1}\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{3}$$

Term 3: tan^-1(sin(-π/2)). Since sin(-π/2) = -1, we get: $$\tan^{-1}(-1) = -\frac{\pi}{4}$$

Term 4: tan^-1(tan(2π/3)). Since 2π/3 is outside the principal range (-π/2, π/2), we use tan(π - θ) = -tan(θ): $$\tan^{-1}\left(\tan\left(\pi - \frac{\pi}{3}\right)\right) = \tan^{-1}\left(-\tan\frac{\pi}{3}\right) = -\frac{\pi}{3}$$

Step 2: Sum the results

Combine the values obtained: $$-\frac{\pi}{6} + \frac{\pi}{3} - \frac{\pi}{4} - \frac{\pi}{3}$$

Simplify the expression: $$-\frac{\pi}{6} - \frac{\pi}{4} = \frac{-2\pi - 3\pi}{12} = -\frac{5\pi}{12}$$