If \(\frac{1 + \cos A}{1 - \cos A} = \left(\frac{m}{n}\right)^2\), then find the value of \(\operatorname{cosec} A - \cot A\).
\(\frac{n}{m}\)
Write the required expression in terms of cos A and sin A: \(\operatorname{cosec} A - \cot A = \frac{1}{\sin A} - \frac{\cos A}{\sin A} = \frac{1 - \cos A}{\sin A}\).
Multiply numerator and denominator by (1 + cos A): \(\frac{1 - \cos A}{\sin A} = \frac{1 - \cos^2 A}{\sin A (1 + \cos A)} = \frac{\sin A}{1 + \cos A}\). So \((\operatorname{cosec} A - \cot A)^2 = \frac{1 - \cos A}{\sin A} \times \frac{\sin A}{1 + \cos A} = \frac{1 - \cos A}{1 + \cos A}\).
Given \(\frac{1 + \cos A}{1 - \cos A} = \frac{m^2}{n^2}\), the reciprocal gives \((\operatorname{cosec} A - \cot A)^2 = \frac{n^2}{m^2}\), so \(\operatorname{cosec} A - \cot A = \frac{n}{m}\) (taking the positive value).
Check: take A = 60°. Then \(\frac{1 + \frac{1}{2}}{1 - \frac{1}{2}} = 3\), so m/n = √3, and \(\operatorname{cosec} 60^\circ - \cot 60^\circ = \frac{2}{\sqrt{3}} - \frac{1}{\sqrt{3}} = \frac{1}{\sqrt{3}}\), which equals n/m.
Hence, \(\frac{n}{m}\).
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