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In Triangle Abc, Prove the Following: a + B C = Cos ( a − B 2 ) Sin C 2 - Mathematics

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प्रश्न

In triangle ABC, prove the following: 

\[\frac{a + b}{c} = \frac{\cos \left( \frac{A - B}{2} \right)}{\sin \frac{C}{2}}\]

 

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उत्तर

Let 

\[\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k\]    ...(1) 

Then,
Consider the LHS of the equation

\[\frac{a + b}{c} = \frac{\cos \left( \frac{A - B}{2} \right)}{\sin \frac{C}{2}}\]

\[LHS = \frac{a + b}{c}\]
\[ = \frac{k\sin A + k\sin B}{k\sin C} \left( using \left( 1 \right) \right)\]
\[ = \frac{2\sin\left( \frac{A + B}{2} \right)\cos\left( \frac{A - B}{2} \right)}{2\sin\frac{C}{2}\cos\frac{C}{2}}\]
\[ = \frac{\sin\left( \frac{A + B}{2} \right)\cos\left( \frac{A - B}{2} \right)}{\sin\frac{C}{2}\cos\left( \frac{\pi - \left( A + B \right)}{2} \right)} \left( \because A + B + C = \pi \right)\]
\[ = \frac{\sin\left( \frac{A + B}{2} \right)\cos\left( \frac{A - B}{2} \right)}{\sin\frac{C}{2}\sin\left( \frac{A + B}{2} \right)}\]
\[ = \frac{\cos\left( \frac{A - B}{2} \right)}{\sin\frac{C}{2}} = RHS \]
\[\text{ Hence proved } .\]

 

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Sine and Cosine Formulae and Their Applications
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पाठ 10: Sine and cosine formulae and their applications - Exercise 10.1 [पृष्ठ १३]

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आरडी शर्मा Mathematics [English] Class 11
पाठ 10 Sine and cosine formulae and their applications
Exercise 10.1 | Q 8 | पृष्ठ १३

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