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Prove That: Cos a + Cos B Cos B − Cos a = Cot ( a + B 2 ) Cot ( a − B 2 )

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प्रश्न

Prove that:

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

Consider LHS: 
\[ \frac{\cos A + cos B}{\cos B - \cos A}\]
\[ = \frac{2\cos \left( \frac{A - B}{2} \right) \cos \left( \frac{A + B}{2} \right)}{2\sin \left( \frac{A + B}{2} \right) \sin \left( \frac{A - B}{2} \right)} \left[ \because \cos A + \cos B = 2\cos \left( \frac{A - B}{2} \right) \cos \left( \frac{A + B}{2} \right) and \cos A - \cos B = 2\sin \left( \frac{A + B}{2} \right) cos \left( \frac{B - A}{2} \right) \right]\]
\[ = \frac{\cos \left( \frac{A - B}{2} \right) \cos \left( \frac{A + B}{2} \right)}{\sin \left( \frac{A + B}{2} \right) sin \left( \frac{A - B}{2} \right)}\]
\[ = \cot\left( \frac{A + B}{2} \right)\cot\left( \frac{A - B}{2} \right)\]
=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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आर.डी. शर्मा Mathematics [English] Class 11
अध्याय 8 Transformation formulae
Exercise 8.2 | Q 7.5 | पृष्ठ १८

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