If \(\cot\theta + \cos\theta = m\), \(\cot\theta - \cos\theta = n\), where \(0 < \theta < \dfrac{\pi}{2}\), then what is \(\dfrac{m^{2}-n^{2}}{\sqrt{mn}}\) equal to?
\(4\)
Adding and subtracting the two given relations, \(m+n = 2\cot\theta\) and \(m-n = 2\cos\theta\). So \(m^{2}-n^{2} = (m+n)(m-n) = 4\cot\theta\cos\theta\). Also \(mn = \cot^{2}\theta-\cos^{2}\theta = \dfrac{\cos^{2}\theta(1-\sin^{2}\theta)}{\sin^{2}\theta} = \dfrac{\cos^{4}\theta}{\sin^{2}\theta}\), so \(\sqrt{mn} = \dfrac{\cos^{2}\theta}{\sin\theta}\) (positive since \(0<\theta<\pi/2\)). Hence \(\dfrac{m^{2}-n^{2}}{\sqrt{mn}} = 4\cot\theta\cos\theta \cdot \dfrac{\sin\theta}{\cos^{2}\theta} = 4\).
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