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In Triangle Abc, Prove the Following: √ Sin a − √ Sin B √ Sin a + √ Sin B = a + B − 2 √ a B a − B - Mathematics

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Question

In triangle ABC, prove the following: 

\[\frac{\sqrt{\sin A} - \sqrt{\sin B}}{\sqrt{\sin A} + \sqrt{\sin B}} = \frac{a + b - 2\sqrt{ab}}{a - b}\]

 

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Solution

Consider the LHS of the equation

\[\frac{\sqrt{\sin A} - \sqrt{\sin B}}{\sqrt{\sin A} + \sqrt{\sin B}} = \frac{a + b - 2\sqrt{ab}}{a - b}\]

\[LHS = \frac{\sqrt{\sin A} - \sqrt{\sin B}}{\sqrt{\sin A} + \sqrt{sin B}}\]
\[ = \frac{\sqrt{\sin A} - \sqrt{\sin B}}{\sqrt{\sin A} + \sqrt{sin B}} \times \frac{\sqrt{\sin A} - \sqrt{\sin B}}{\sqrt{\sin A} - \sqrt{\sin B}}\]
\[ = \frac{\sin A + \sin B - \left( 2 \times \sqrt{\sin A\sin B} \right)}{\sin A - \sin B}\]

Let \[\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k\]

Then,

\[LHS=\frac{\frac{a}{k} + \frac{b}{k} - 2 \times \sqrt{\frac{a}{k}\frac{b}{k}}}{\frac{a}{k} - \frac{b}{k}} \]

\[ = \frac{\frac{1}{k}\left( a + b - 2 \times \sqrt{ab} \right)}{\frac{1}{k}\left( a - b \right)}\]

\[ =\frac{a + b - 2\sqrt{ab}}{a - b}=RHS\]

\[\text{ Hence proved } \]

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Sine and Cosine Formulae and Their Applications
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Chapter 10: Sine and cosine formulae and their applications - Exercise 10.1 [Page 13]

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RD Sharma Mathematics [English] Class 11
Chapter 10 Sine and cosine formulae and their applications
Exercise 10.1 | Q 13 | Page 13

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