Question

If $\alpha ,\beta ,\gamma$ are lengths of the altitudes of a triangle ABC then $\frac{1}{{\alpha }^2}+\frac{1}{{\beta }^2}+\frac{1}{{\gamma }^2}$ is equal to

• A. $\frac{cotA+cotB+cotC}{\Delta }$
• B. $\frac{\Delta }{cotA+cotB+cotC}$
• C. $\Delta (cotA+cotB+cotC)$
• D. none of these

Answer 1

The correct option is A $\frac{cotA+cotB+cotC}{\Delta }$

we know in $\Delta ABC,$ $\Delta =\frac{1}{2}a\alpha =\frac{1}{2}b\beta =\frac{1}{2}c\gamma$

$\Rightarrow \frac{1}{{\alpha }^2}+\frac{1}{{\beta }^2}+\frac{1}{{\gamma }^2}=\frac{a^2+b^2+c^2}{4{\Delta }^2}$
and $cotA+cotB+cotC$

$=\frac{2R}{2abc}[b^2+c^2-a^2+c^2+a^2-b^2+a^2+b^2-c^2]$

$=\frac{R(a^2+b^2+c^2)}{abc}=\frac{a^2+b^2+c^2}{4\Delta }$

Hence $\frac{1}{{\alpha }^2}+\frac{1}{{\beta }^2}+\frac{1}{{\gamma }^2}=\frac{cotA+cotB+cotC}{\Delta }$