Question

The base of a triangle is divided into three equal parts. If $\alpha ,\beta ,\gamma$ be the angle subtended by these parts of vertex, prove that: $(\cot \alpha +\cot \beta )(\cot \beta +\cot \gamma )=4cose{c}^2\beta$

Answer 1

Let point $D$ and $E$ divide the base $BC$ into three equal parts i.e., $BD=DE=DC=d$ (let)
Let $\alpha ,\beta$ and $\gamma$ be the angles subtended by $BD,DE$ and $EC$ respectively at their opposite vertex.
Now in $△ABC$
$\because BE:EC=2d:d=2:1$
$\therefore$ from $m-n$ rule, we get
$(2+1)\cot \theta =2\cot (\alpha +\beta )-\cot \gamma$
$\Rightarrow 3\cot \theta =2\cot (\alpha +\beta )-\cot \gamma$ ...(i)
Again since in $△ADC,DE:EC=x:x=1:1$
If we apply $m-n$ rule in $△ADC$ we get
$(1+1)\cot \theta =1-\cot \beta -1\cot \gamma$
$\Rightarrow 2\cot \theta =\cot \beta -\cot \gamma$ ...(ii)
From (i) and (ii), we get
$\frac{3\cot \theta }{2\cot \theta }=\frac{2\cot (\alpha +\beta )-\cot \gamma }{\cot \beta -\cot \gamma }$
$\Rightarrow 3\cot \beta -3\cot \gamma =4\cot (\alpha +\beta )-2\cot \gamma$
$\Rightarrow 3\cot \beta -\cot \gamma =4\cot (\alpha +\beta )\Rightarrow 3\cot \beta -\cot \gamma =4\left\{\frac{\cot \alpha \cot \beta -1}{\cot \beta +\cot \alpha }\right\}$
$\Rightarrow 3{\cot }^2\beta +3\cot \alpha \cot \beta -\cot \beta \cot \gamma -\cot \alpha \cot \gamma =4\cot \alpha \cot \beta -4$
$4+3{\cot }^2\beta =\cot \alpha \cot \beta +\cot \beta \cot \gamma +\cot \alpha \cot \gamma$
$\Rightarrow 4+4{\cot }^2\beta =\cot \alpha \cot \beta +\cot \alpha \cot \gamma +\cot \beta \cot \gamma +{\cot }^2\beta$
$\Rightarrow 4(1+\frac{1}{{\tan }^2\beta })=(\frac{1}{\tan \alpha }+\frac{1}{\tan \beta })(\frac{1}{\tan \beta }+\frac{1}{\tan \gamma })$
$4{\csc }^2\beta =(\cot \alpha +\cot \beta )(\cot \beta +\cot \gamma )$