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प्रश्न
Prove that:
उत्तर
Consider LHS:
\[ \frac{\sin A - \sin B}{\cos A + \cos B}\]
\[ = \frac{2\sin \left( \frac{A - B}{2} \right) \cos \left( \frac{A + B}{2} \right)}{2\cos \left( \frac{A + B}{2} \right) \cos \left( \frac{A - B}{2} \right)} \left[ \because \sin A - \sin B = 2\sin \left( \frac{A - B}{2} \right) \cos \left( \frac{A + B}{2} \right) and \cos A + \cos B = 2\cos \left( \frac{A + B}{2} \right) \cos \left( \frac{A - B}{2} \right) \right]\]
\[ = \frac{\sin \left( \frac{A - B}{2} \right) \cos \left( \frac{A + B}{2} \right)}{\cos \left( \frac{A + B}{2} \right) \cos \left( \frac{A - B}{2} \right)}\]
\[ = \tan\left( \frac{A - B}{2} \right)\]
= RHS
Hence, LHS = RHS.
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