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Question
If A, B, C are in A.P., then \[\frac{\sin A - \sin C}{\cos C - \cos A}\]=
Options
tan B
cot B
tan 2 B
None of these
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Solution
cot B
Since A,B and C are in A.P,
B - A = C - B
or, 2B = A + C
\[\frac{\sin A - \sin C}{\cos C - \cos A}\]
\[ = \frac{2\sin\left( \frac{A - C}{2} \right)\cos\left( \frac{A + C}{2} \right)}{- 2\sin\left( \frac{C + A}{2} \right)\sin\left( \frac{C - A}{2} \right)} \left[ \because \sin A - \sin B = 2\sin\left( \frac{A - B}{2} \right)\cos\left( \frac{A + B}{2} \right) \text{ and }\cos A - \cos B = - 2\sin\left( \frac{A + B}{2} \right)\cos\left( \frac{A - B}{2} \right) \right]\]
\[ = \frac{\sin\left( \frac{A - C}{2} \right)\cos\left( \frac{A + C}{2} \right)}{- \sin\left( \frac{A + C}{2} \right)\sin\left( \frac{C - A}{2} \right)}\]
\[= \frac{\sin\left( \frac{A - C}{2} \right)\cos\left( \frac{A + C}{2} \right)}{\sin\left( \frac{A + C}{2} \right)\sin\left( \frac{A - C}{2} \right)}\]
\[ = \frac{\cos\left( \frac{A + C}{2} \right)}{\sin\left( \frac{A + C}{2} \right)}\]
\[ = \frac{\cos B}{\sin B}\]
\[ = \cot B\]
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