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

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प्रश्न

Prove that:

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

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
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 8: Transformation formulae - Exercise 8.2 [पृष्ठ १८]

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आर.डी. शर्मा Mathematics [English] Class 11
अध्याय 8 Transformation formulae
Exercise 8.2 | Q 7.3 | पृष्ठ १८

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