Let Delta denote the area of a triangle ABC. | vedclass

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(B) The area of $	riangle ABC$ is given by $\Delta = \frac{1}{2} a \alpha = \frac{1}{2} b \beta = \frac{1}{2} c \gamma$. Thus,$\alpha = \frac{2\Delta

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$\frac{1}{\Delta}(\cot A+\cot B+\cot C)$

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(B) The area of $	riangle ABC$ is given by $\Delta = \frac{1}{2} a \alpha = \frac{1}{2} b \beta = \frac{1}{2} c \gamma$. Thus,$\alpha = \frac{2\Delta}{a}$,$\beta = \frac{2\Delta}{b}$,and $\gamma = \frac{2\Delta}{c}$. Substituting these into the expression: $\alpha^{-2} + \beta^{-2} + \gamma^{-2} = \frac{a^2}{4\Delta^2} + \frac{b^2}{4\Delta^2} + \frac{c^2}{4\Delta^2} = \frac{a^2+b^2+c^2}{4\Delta^2}$. Using the identity $\cot A = \frac{b^2+c^2-a^2}{4\Delta}$,we have $b^2+c^2-a^2 = 4\Delta \cot A$. Summing these for $A, B, C$: $(b^2+c^2-a^2) + (c^2+a^2-b^2) + (a^2+b^2-c^2) = a^2+b^2+c^2 = 4\Delta(\cot A + \cot B + \cot C)$. Therefore,$\frac{a^2+b^2+c^2}{4\Delta^2} = \frac{4\Delta(\cot A + \cot B + \cot C)}{4\Delta^2} = \frac{1}{\Delta}(\cot A + \cot B + \cot C)$.

Let $\Delta$ denote the area of a $\triangle ABC$. If $\alpha, \beta, \gamma$ are the lengths of the altitudes of the $\triangle ABC$,then $\alpha^{-2}+\beta^{-2}+\gamma^{-2}=$

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