If \(\cot(A - B) = \sqrt{3}\), then find the value of \(\frac{\cot B - \cot A}{\cot A\cot B + 1} + \frac{\tan A - \tan B}{1 + \tan A\tan B}\).
\(\frac{2\sqrt{3}}{3}\)
Recall \(\cot(A - B) = \frac{\cot A\cot B + 1}{\cot B - \cot A}\), so the first term \(\frac{\cot B - \cot A}{\cot A\cot B + 1} = \frac{1}{\cot(A - B)} = \tan(A - B)\)
Also \(\frac{\tan A - \tan B}{1 + \tan A\tan B} = \tan(A - B)\)
Given \(\cot(A - B) = \sqrt{3}\), so \(\tan(A - B) = \frac{1}{\sqrt{3}}\)
Sum = \(\frac{1}{\sqrt{3}} + \frac{1}{\sqrt{3}} = \frac{2}{\sqrt{3}} = \frac{2\sqrt{3}}{3}\)
Hence, the answer is \(\frac{2\sqrt{3}}{3}\).
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