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Question
Prove that:
\[\frac{1}{\sin \left( x - a \right) \sin \left( x - b \right)} = \frac{\cot \left( x - a \right) - \cot \left( x - b \right)}{\sin \left( a - b \right)}\]
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Solution
\[\text{ RHS }\hspace{0.167em} = \frac{\cot\left( x - a \right) - \cot(x - b)}{\sin(a - b)}\]
\[ = \frac{\frac{\cos(x - a)}{\sin(x - a)} - \frac{\cos(x - b)}{\sin(x - b)}}{\sin(a - b)}\]
\[ = \frac{\sin(x - b) \cos(x - a) - \sin(x - a) \cos(x - b)}{\sin(x - a) \sin(x - b) \sin(a - b)}\]
\[ = \frac{\sin(x - b - x + a)}{\sin(x - a) \sin(x - b) \sin(a - b)}\]
\[ = \frac{\sin(a - b)}{\sin(x - a) \sin(x - b) \sin(a - b)}\]
\[ = \frac{1}{\sin(x - a)\sin(x - b)} \]
= LHS
Hence proved.
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