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Let $\triangle$ denote the area of a $\triangle A B C$. If $\alpha, \beta, \gamma$ are the lengths of the altitudes of the $\triangle A B C$, then $\alpha^{-2}+\beta^{-2}+\gamma^{-2}=$
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Verified Answer
The correct answer is:
$\frac{1}{\Delta}(\cot A+\cot B+\cot C)$
$$
\begin{aligned}
& \text { Since, } \alpha^{-2}+\beta^{-2}+\gamma^{-2}=\frac{a^2}{4 \Delta^2}+\frac{b^2}{4 \Delta^2}+\frac{c^2}{4 \Delta^2} \\
& =\frac{a^2+b^2+c^2}{4 \Delta^2} \\
& =\frac{\left(a^2+b^2-c^2\right)+\left(b^2+c^2-a^2\right)+\left(c^2+a^2-b^2\right)}{4\left(\frac{a b c}{4 R}\right)^2} \\
& =\frac{2 a b \cos C+2 b c \cos A+2 a c \cos B}{\frac{a^2 b^2 c^2}{4 R^2}} \\
& =\frac{\cos C}{\left(\frac{a b c}{4 R}\right) \frac{c}{2 R}+\frac{\cos A}{\left(\frac{a b c}{4 R}\right) \frac{a}{2 R}}+\frac{\cos B}{\left(\frac{a b c}{4 R}\right) \frac{b}{2 R}}} \\
& =\frac{1}{\Delta}(\cot A+\cot B+\cot C)
\end{aligned}
$$
Hence, option (b) is correct.
\begin{aligned}
& \text { Since, } \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} \\
& =\frac{\left(a^2+b^2-c^2\right)+\left(b^2+c^2-a^2\right)+\left(c^2+a^2-b^2\right)}{4\left(\frac{a b c}{4 R}\right)^2} \\
& =\frac{2 a b \cos C+2 b c \cos A+2 a c \cos B}{\frac{a^2 b^2 c^2}{4 R^2}} \\
& =\frac{\cos C}{\left(\frac{a b c}{4 R}\right) \frac{c}{2 R}+\frac{\cos A}{\left(\frac{a b c}{4 R}\right) \frac{a}{2 R}}+\frac{\cos B}{\left(\frac{a b c}{4 R}\right) \frac{b}{2 R}}} \\
& =\frac{1}{\Delta}(\cot A+\cot B+\cot C)
\end{aligned}
$$
Hence, option (b) is correct.
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