$y\left( \alpha \right) = \sqrt {2\left( {{{\tan \alpha + \cot \alpha } \over {1 + {{\tan }^2}\alpha }}} \right) + {1 \over {{{\sin }^2}\alpha }}}$ = $\sqrt {2\left( {{{1 + {{\tan }^2}\alpha } \over {\tan \alpha \left( {1 + {{\tan }^2}\alpha } \right)}}} \right) + {1 \over {{{\sin }^2}\alpha }}}$ = $\sqrt {2\left( {{{\cos \alpha } \over {\sin \alpha }}} \right) + {1 \over {{{\sin }^2}\alpha }}}$ = $\sqrt {{{\sin 2\alpha + 1} \over {{{\sin }^2}\alpha }}}$ = ${{\left| {\sin \alpha + \cos \alpha } \right|} \over {\left| {\sin \alpha } \right|}}$ 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$\alpha$ $\therefore$ ${{dy} \over {d\alpha }}$ = cosec 2 $\alpha$ So ${{{dy} \over {d\alpha }}}$ at $\alpha$ = ${{5\pi } \over 6}$, = cosec 2 ${{5\pi } \over 6}$ = 4