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If A, B, C Are in A.P., Then Sin a − Sin C Cos C − Cos a = - Mathematics

MCQ
Sum

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
\[\frac{\sin A - \sin C}{\cos C - \cos A}\]
\[ = \frac{2\sin\left( \frac{A - C}{2} \right)\cos\left( \frac{A + C}{2} \right)}{- 2\sin\left( \frac{C + A}{2} \right)\sin\left( \frac{C - A}{2} \right)} \left[ \because \sin A - \sin B = 2\sin\left( \frac{A - B}{2} \right)\cos\left( \frac{A + B}{2} \right) \text{ and }\cos A - \cos B = - 2\sin\left( \frac{A + B}{2} \right)\cos\left( \frac{A - B}{2} \right) \right]\]
\[ = \frac{\sin\left( \frac{A - C}{2} \right)\cos\left( \frac{A + C}{2} \right)}{- \sin\left( \frac{A + C}{2} \right)\sin\left( \frac{C - A}{2} \right)}\]
\[= \frac{\sin\left( \frac{A - C}{2} \right)\cos\left( \frac{A + C}{2} \right)}{\sin\left( \frac{A + C}{2} \right)\sin\left( \frac{A - C}{2} \right)}\]
\[ = \frac{\cos\left( \frac{A + C}{2} \right)}{\sin\left( \frac{A + C}{2} \right)}\]
\[ = \frac{\cos B}{\sin B}\]
\[ = \cot B\]

Concept: Transformation Formulae
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APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 8 Transformation formulae
Q 11 | Page 21
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