Prove that: (B^2-c^2)cotA+(c^2-a^2)cotB+(a^2-b^2)cotC=0
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Posted by Onkar Bansal 7 years, 10 months ago
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Samir Kumar Basu 7 years, 10 months ago
(b2−c2).cotA+(c2−a2).cotB+(a2−b2).cotC=(b2−c2)R(b2+c2−a2)abc+(c2−a2).R(c2+a2−b2)abc+(a2−b2).R(a2+b2−c2)abc=Rabc[b4−c4−a2b2+c2a2+c4−a4−b2c2+a2b2+a4−b4−c2a2+b2c2]=0[Formulaused:cotA=R(b2+c2−a2)abc,cotB=R(c2+a2−b2)abc,cotC=R(a2+b2−c2)abc]
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