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प्रश्न
In triangle ABC, prove the following:
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उत्तर
Consider the LHS of the equation
\[LHS = \frac{\sqrt{\sin A} - \sqrt{\sin B}}{\sqrt{\sin A} + \sqrt{sin B}}\]
\[ = \frac{\sqrt{\sin A} - \sqrt{\sin B}}{\sqrt{\sin A} + \sqrt{sin B}} \times \frac{\sqrt{\sin A} - \sqrt{\sin B}}{\sqrt{\sin A} - \sqrt{\sin B}}\]
\[ = \frac{\sin A + \sin B - \left( 2 \times \sqrt{\sin A\sin B} \right)}{\sin A - \sin B}\]
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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