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Question
Prove that:
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Solution
\[\text{ RHS }= \frac{\tan(x - b) - \tan(x - a)}{\sin(a - b)}\]
\[ = \frac{\frac{\sin(x - b)}{\cos(x - b)} - \frac{\sin(x - a)}{\cos(x - a)}}{\sin(a - b)}\]
\[ = \frac{\sin(x - b) \cos(x - a) - \sin(x - a) \cos(x - b)}{\sin(a - b) \cos(x - a) \cos(x - b)}\]
\[ = \frac{\sin(x - b - x + a)}{\sin(a - b) \cos(x - a) \cos(x - b)} ( \text{ Using }\sin(A - B) = \sin A\cos B - \cos A\sin B)\]
\[ = \frac{\sin(a - b)}{\sin(a - b) \cos(x - a) \cos(x - b)}\]
\[ = \frac{1}{\cos(x - a) \cos(x - b)} \]
= LHS
Hence proved.
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