Official solution
\[y\left( \alpha \right) = \sqrt {2\left( {{{\tan \alpha + \cot \alpha } \over {1 + {{\tan }^2}\alpha }}} \right) + {1 \over {{{\sin }^2}\alpha }}}\]
= \[\sqrt {2\left( {{{1 + {{\tan }^2}\alpha } \over {\tan \alpha \left( {1 + {{\tan }^2}\alpha } \right)}}} \right) + {1 \over {{{\sin }^2}\alpha }}} \]
= \[\sqrt {2\left( {{{\cos \alpha } \over {\sin \alpha }}} \right) + {1 \over {{{\sin }^2}\alpha }}} \]
= \[\sqrt {{{\sin 2\alpha + 1} \over {{{\sin }^2}\alpha }}} \]
= \[{{\left| {\sin \alpha + \cos \alpha } \right|} \over {\left| {\sin \alpha } \right|}}\]
At \[\alpha \] = \[{{5\pi } \over 6}\]
\[{\left| {\sin \alpha + \cos \alpha } \right|}\] = -(sin\[\alpha \] + cos\[\alpha \])
and |sin\[\alpha \]| = sin\[\alpha \]
\[ \therefore \] y(\[\alpha \]) = \[{{ - \left( {\sin \alpha + \cos \alpha } \right)} \over {\sin \alpha }}\]
= -1 - cot\[\alpha \]
\[ \therefore \] \[{{dy} \over {d\alpha }}\] = cosec2\[\alpha \]
So \[{{{dy} \over {d\alpha }}}\] at \[\alpha \] = \[{{5\pi } \over 6}\],
= cosec2\[{{5\pi } \over 6}\] = 4