Official solution
Since, x2 + 3y2 = 9
\[ \Rightarrow \] 2x + 6y \[{{dy} \over {dx}}\] = 0
\[ \Rightarrow \] \[{{dy} \over {dx}}\] = \[{{ - x} \over {3y}}\]
Slope of normal is - \[{{dx} \over {dy}}\] = \[{{3y} \over x}\]
\[ \Rightarrow \] \[{\left( { - {{dx} \over {dy}}} \right)_{\left( {3\cos \theta ,\sqrt 3 \sin \theta } \right)}}\]
= \[{{3\sqrt 3 \sin \theta } \over {3\cos \theta }}\] = \[\sqrt 3 \tan \theta \] = m1
\& \[{\left( { - {{dx} \over {dy}}} \right)_{\left( { - 3\sin \theta ,\sqrt 3 \cos \theta } \right)}}\]
= \[{{3\sqrt 3 \cos \theta } \over { - 3\sin \theta }}\] = \[ - \sqrt 3 \cot \theta \] = m2
As, \[\beta \] is the angle between the normals to the given ellipse then
tan\[\beta \] = \[\left| {{{{m_1} - {m_2}} \over {1 + {m_1}{m_2}}}} \right|\]
= \[\left| {{{\sqrt 3 \tan \theta + \sqrt 3 \cot \theta } \over {1 - 3\tan \theta \cot \theta }}} \right|\] = \[\left| {{{\sqrt 3 \tan \theta + \sqrt 3 \cot \theta } \over {1 - 3}}} \right|\]
So, tan \[\beta \] = \[{{\sqrt 3 } \over 2}\] \[\left| {\tan \theta + \cot \theta } \right|\]
\[ \Rightarrow \] \[{1 \over {\cot \beta }} = {{\sqrt 3 } \over 2}\left| {{{\sin \theta } \over {\cos \theta }} + {{\cos \theta } \over {\sin \theta }}} \right|\]
\[ \Rightarrow \] \[{1 \over {\cot \beta }} = {{\sqrt 3 } \over 2}\] \[\left| {{1 \over {\sin \theta \cos \theta }}} \right|\]
\[ \Rightarrow \] \[{1 \over {\cot \beta }} = {{\sqrt 3 } \over {\sin 2\theta }}\]
\[ \Rightarrow \] \[{{2\cot \beta } \over {\sin 2\theta }}\] = \[{2 \over {\sqrt 3 }}\]