COS \frac{B+C}{2}\sin \frac{B-C}{2}\cot \frac{A}{2}\] \[=2k\sin \frac{A}{2}\,.\,\,\sin \frac{B-C}{2}\TEXT{ }\text{. }\frac{\cos \frac{A}{2}}{\sin \frac{A}{2}}\] \[=2k\sin \left( \frac{B-C}{2} \RIGHT)\sin \left( \frac{B+C}{2} \right)=2k\left( {{\sin }^{2}}\frac{B}{2}-{{\sin }^{2}}\frac{C}{2} \right)\] or we GET L.H.S. =\[\Sigma 2k\left( {{\sin }^{2}}\frac{B}{2}-{{\sin }^{2}}\frac{C}{2} \right)=0\].

"> COS \frac{B+C}{2}\sin \frac{B-C}{2}\cot \frac{A}{2}\] \[=2k\sin \frac{A}{2}\,.\,\,\sin \frac{B-C}{2}\TEXT{ }\text{. }\frac{\cos \frac{A}{2}}{\sin \frac{A}{2}}\] \[=2k\sin \left( \frac{B-C}{2} \RIGHT)\sin \left( \frac{B+C}{2} \right)=2k\left( {{\sin }^{2}}\frac{B}{2}-{{\sin }^{2}}\frac{C}{2} \right)\] or we GET L.H.S. =\[\Sigma 2k\left( {{\sin }^{2}}\frac{B}{2}-{{\sin }^{2}}\frac{C}{2} \right)=0\].

">
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In \[\Delta ABC\], \[(b-c)\cot \frac{A}{2}+(c-a)\cot \frac{B}{2}+(a-b)\]\[\cot \frac{C}{2}\] is equal to [WB JEE 1989]

Joint Entrance Exam - Main (JEE Main) Mathematics in Joint Entrance Exam - Main (JEE Main) 10 months ago

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\[(b-c)\cot \frac{A}{2}=k(\sin B-\sin C)\cot \frac{A}{2}\] \[=2k\COS \frac{B+C}{2}\sin \frac{B-C}{2}\cot \frac{A}{2}\] \[=2k\sin \frac{A}{2}\,.\,\,\sin \frac{B-C}{2}\TEXT{ }\text{. }\frac{\cos \frac{A}{2}}{\sin \frac{A}{2}}\] \[=2k\sin \left( \frac{B-C}{2} \RIGHT)\sin \left( \frac{B+C}{2} \right)=2k\left( {{\sin }^{2}}\frac{B}{2}-{{\sin }^{2}}\frac{C}{2} \right)\] or we GET L.H.S. =\[\Sigma 2k\left( {{\sin }^{2}}\frac{B}{2}-{{\sin }^{2}}\frac{C}{2} \right)=0\].

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Posted on 17 Sep 2024, this text provides information on Joint Entrance Exam - Main (JEE Main) related to Mathematics in Joint Entrance Exam - Main (JEE Main). Please note that while accuracy is prioritized, the data presented might not be entirely correct or up-to-date. This information is offered for general knowledge and informational purposes only, and should not be considered as a substitute for professional advice.

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