If \(\dfrac{1}{\operatorname{cosec}\theta+\cot\theta} - \dfrac{1}{\sin\theta} = p\), then what is \(\dfrac{1}{\sin\theta} - \dfrac{1}{\operatorname{cosec}\theta-\cot\theta}\) equal to?
\(p\)
With \(s=\sin\theta,\ c=\cos\theta\): \(\operatorname{cosec}\theta+\cot\theta = \dfrac{1+c}{s}\), so \(\dfrac{1}{\operatorname{cosec}\theta+\cot\theta}-\dfrac{1}{s} = \dfrac{s}{1+c}-\dfrac{1}{s} = \dfrac{s^{2}-(1+c)}{s(1+c)} = \dfrac{-c(1+c)}{s(1+c)} = -\cot\theta = p\). Similarly \(\operatorname{cosec}\theta-\cot\theta = \dfrac{1-c}{s}\), so \(\dfrac{1}{s}-\dfrac{1}{\operatorname{cosec}\theta-\cot\theta} = \dfrac{1}{s}-\dfrac{s}{1-c} = \dfrac{(1-c)-s^{2}}{s(1-c)} = \dfrac{c^{2}-c}{s(1-c)} = -\cot\theta\). Since both equal \(-\cot\theta\), the required expression equals \(p\).
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