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Prove That: Sin a − Sin B Cos a + Cos B = Tan a − B 2

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Question

Prove that:

\[\frac{\sin A - \sin B}{\cos A + \cos B} = \tan\frac{A - B}{2}\]

Sum

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Solution

Consider LHS:
\[ \frac{\sin A - \sin B}{\cos A + \cos B}\]
\[ = \frac{2\sin \left( \frac{A - B}{2} \right) \cos \left( \frac{A + B}{2} \right)}{2\cos \left( \frac{A + B}{2} \right) \cos \left( \frac{A - B}{2} \right)} \left[ \because \sin A - \sin B = 2\sin \left( \frac{A - B}{2} \right) \cos \left( \frac{A + B}{2} \right) and \cos A + \cos B = 2\cos \left( \frac{A + B}{2} \right) \cos \left( \frac{A - B}{2} \right) \right]\]
\[ = \frac{\sin \left( \frac{A - B}{2} \right) \cos \left( \frac{A + B}{2} \right)}{\cos \left( \frac{A + B}{2} \right) \cos \left( \frac{A - B}{2} \right)}\]
\[ = \tan\left( \frac{A - B}{2} \right)\]
= RHS
Hence, LHS = RHS.

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Transformation Formulae

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Q 7.3 Q 7.2 Q 7.4

Chapter 8: Transformation formulae - Exercise 8.2 [Page 18]

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R.D. Sharma Mathematics [English] Class 11

Chapter 8 Transformation formulae
Exercise 8.2 | Q 7.3 | Page 18

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