Let \(t = \tan\theta\) (with \(t > 0\) for \(0^\circ < \theta < 90^\circ\)). Using the angle addition formulas: \[ \tan(\theta \pm 60^\circ) = \frac{t \pm \sqrt{3}}{1 \mp \sqrt{3}t}. \] Substitute into the equation \[ \tan(\theta + 60^\circ) + \tan(\theta - 60^\circ) = \cot\theta = \frac{1}{t}, \] to get \[ \frac{t + \sqrt{3}}{1 - \sqrt{3}t} + \frac{t - \sqrt{3}}{1 + \sqrt{3}t} = \frac{1}{t}. \]

Combine the fractions on the left: \[ \frac{(t + \sqrt{3})(1 + \sqrt{3}t) + (t - \sqrt{3})(1 - \sqrt{3}t)}{(1 - \sqrt{3}t)(1 + \sqrt{3}t)} = \frac{8t}{1 - 3t^2}. \] So the equation becomes \[ \frac{8t}{1 - 3t^2} = \frac{1}{t}. \]

Multiply through by \(t(1 - 3t^2)\): \[ 8t^2 = 1 - 3t^2 \quad \Rightarrow \quad 11t^2 = 1 \quad \Rightarrow \quad t^2 = \frac{1}{11}. \] Since \(t > 0\), \[ t = \frac{1}{\sqrt{11}}. \]

Hence, \[ \theta = \arctan\!\frac{1}{\sqrt{11}} \approx 16.8^\circ, \] which is the only solution for \(0^\circ < \theta < 90^\circ\).