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Prove That: Sin a + Sin B Sin a − Sin B = Tan ( a + B 2 ) Cot ( a − B 2 ) - Mathematics

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प्रश्न

Prove that:

\[\frac{\sin A + \sin B}{\sin A - \sin B} = \tan \left( \frac{A + B}{2} \right) \cot \left( \frac{A - B}{2} \right)\]
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उत्तर

Consider LHS:
\[ \frac{\sin A + \sin B}{\sin A - \sin B}\]
\[ = \frac{2\sin \left( \frac{A + B}{2} \right) \cos \left( \frac{A - B}{2} \right)}{2\sin \left( \frac{A + B}{2} \right) \cos \left( \frac{A - B}{2} \right)} \left\{ \because \sin A + \sin B = 2\sin \left( \frac{A + B}{2} \right) \cos \left( \frac{A - B}{2} \right), and \sin A - \sin B = 2\sin \left( \frac{A - B}{2} \right) \cos\left( \frac{A + B}{2} \right) \right\}\]
\[ = \frac{\sin \left( \frac{A + B}{2} \right) \cos \left( \frac{A - B}{2} \right)}{\sin \left( \frac{A - B}{2} \right) \cos \left( \frac{A + B}{2} \right)}\]
\[ = \tan \left( \frac{A + B}{2} \right) cot \left( \frac{A - B}{2} \right)\]
 = 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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आरडी शर्मा Mathematics [English] Class 11
अध्याय 8 Transformation formulae
Exercise 8.2 | Q 7.4 | पृष्ठ १८

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