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Prove that:(B^2-c^2)cotA+(c^2-a^2)cotB+(a^2-b^2)cotC=0 |
| Answer» {tex}\\begin{array}{l}\\left(b^2-c^2\\right).cotA+\\left(c^2-a^2\\right).cotB+\\left(a^2-b^2\\right).cotC\\\\=\\left(b^2-c^2\\right)\\frac{R\\left(b^2+c^2-a^2\\right)}{abc}+\\left(c^2-a^2\\right).\\frac{R\\left(c^2+a^2-b^2\\right)}{abc}+\\left(a^2-b^2\\right).\\frac{R\\left(a^2+b^2-c^2\\right)}{abc}\\\\=\\frac R{abc}\\left[b^4-c^4-a^2b^2+c^2a^2+c^4-a^4-b^2c^2+a^2b^2+a^4-b^4-c^2a^2+b^2c^2\\right]\\\\=0\\\\\\;\\lbrack\\;Formula\\;used:\\;\\;\\mathrm{co}t\\;A=\\frac{R\\left(b^2+c^2-a^2\\right)}{abc},\\;cot\\;B=\\frac{R\\left(c^2+a^2-b^2\\right)}{abc},\\;cot\\;C=\\frac{R\\left(a^2+b^2-c^2\\right)}{abc}\\;\\;\\;\\rbrack\\end{array}{/tex} | |