If 0 \leq (\alpha-\beta) \leq (\alpha+\beta) \leq \frac{\pi}{2} where \tan(\alpha+\beta)=\sqrt{3} and \tan(\alpha-\beta)=\frac{1}{\sqrt{3}}, then what is \tan \alpha \cdot \cot 2\beta equal to?

Explanation

From the given equations, \alpha+\beta = 60^\circ and \alpha-\beta = 30^\circ. Solving these gives \alpha = 45^\circ and \beta = 15^\circ. The expression becomes \tan 45^\circ \cdot \cot 30^\circ = 1 \times \sqrt{3} = \sqrt{3}.

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