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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}
Posted by Ranjan Mahto 8 years, 3 months ago
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Onkar Bansal 8 years, 3 months ago
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Samir Kumar Basu 8 years, 3 months ago
{tex}\begin{array}{l}\sin\theta+\cos\theta=\sqrt2\\\Rightarrow\frac1{\sqrt2}\sin\theta+\frac1{\sqrt2}\cos\theta=1\\\Rightarrow\cos\left(\theta-\frac\pi4\right)=\cos\;0\\\Rightarrow\theta-\frac\pi4=2n\pi\;\;\;\;\;\lbrack n\in set\;of\;integers\;\rbrack\\\Rightarrow\theta=2n\pi\;+\frac\pi4\end{array}{/tex}
{tex}\sin\theta=\sin\;\left(2n\pi\;+\frac\pi4\right)\;\;\;=\sin\frac\pi4=\frac1{\sqrt2}{/tex}
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Samir Kumar Basu 8 years, 3 months ago
{tex}\begin{array}{l}LHS=\sin^2A+\sin^2(60^\circ+A)+\sin^2(60^\circ-A)\\=\sin^2A+\frac12\left[1-\cos2(60^\circ+A)\right]+\frac12\left[1-\cos2(60^\circ-A)\right]\\\;\;\;\;\;\;\;\;\lbrack\sin ce\;\;\sin^2\theta=\frac12(1-\cos2\theta)\;\;\rbrack\\=\sin^2A+1-\frac12\left[\cos(120^\circ+2A)+\cos(120^\circ-2A)\right]\\=\sin^2A+1-\frac12\times2\cos\frac{(120^\circ+2A)+(120^\circ-2A)}2.\cos\frac{(120^\circ+2A)-(120^\circ-2A)}2\\\;\;\;\;\;\;\;\;\;\;\;\;\;\;\sin ce\;\;\cos\;C\;+\;\cos\;D=2.\cos\frac{C+D}2.\cos\frac{C-D}2\\=\sin^2A+1-\cos120^\textdegree.\cos2A\\=\sin^2A+1+\frac12.\left(1-2\sin^2A\right)\\=\sin^2A+1+\frac12-\sin^2A\\=\frac32=RHS\end{array}{/tex}
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