Prove that: (B^2-c^2)cotA+(c^2-a^2)cotB+(a^2-b^2)cotC=0
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Samir Kumar Basu 8 years, 3 months ago
{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}
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