If $α, β, γ$ are the altitudes of the triangle ABC from the vertices A,B,C respectively, then the value of $\frac{1}{α^{2}} + \frac{1}{β^{2}} + \frac{1}{γ^{2}}$ is

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$\frac{\cot A + \cot B + \cot C}{Δ}$

$\frac{△}{\cot A + \cot B + \cot C}$

$\frac{\cot A + \cot B + \cot C}{2 Δ}$

$\frac{4}{△}$

answer is A.

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Detailed Solution

$\begin{array}{l} \frac{1}{2} α \cdot a = Δ \Rightarrow \frac{1}{α} = \frac{a}{2 Δ} etc. \\ \frac{1}{α^{2}} + \frac{1}{β^{2}} + \frac{1}{λ^{2}} = \frac{a^{2} + b^{2} + c^{2}}{4 Δ^{2}} - - - (i) \end{array}$

$\cot A + \cot B + \cot C = \frac{\cos A}{\sin A} + \frac{\cos B}{\sin B} + \frac{\cos C}{\sin C}$
$\begin{array}{l} = \frac{b^{2} + c^{2} - a^{2}}{2 b c \cdot a} \cdot 2 R + \frac{c^{2} + a^{2} - b^{2}}{2 a c \cdot b} \cdot 2 R + \frac{a^{2} + b^{2} - c^{2}}{2 a b \cdot c} \cdot 2 R \\ = \frac{R}{a b c} \cdot [a^{2} + b^{2} + c^{2}] \\ = \frac{a b c}{4 Δ \cdot a b c} [a^{2} + b^{2} + c^{2}] = \frac{a^{2} + b^{2} + c^{2}}{4 Δ} - - - (i i) \\ (i), (ii) \Rightarrow \frac{1}{α^{2}} + \frac{1}{β^{2}} + \frac{1}{λ^{2}} = \frac{1}{Δ} (\cot A + \cot B + \cot C) \end{array}$