$2y = {\left( {{{\cot }^{ - 1}}\left( {{{\sqrt 3 \cos x + \sin x} \over {\cos x - \sqrt 3 \sin x}}} \right)} \right)^2}$ $\Rightarrow$ 2y = ${\left( {{{\cot }^{ - 1}}\left( {{{\sqrt 3 + \tan x} \over {1 - \sqrt 3 \tan x}}} \right)} \right)^2}$ $\Rightarrow$ 2y = ${\left( {{{\cot }^{ - 1}}\left( {{{\tan {\pi \over 3} + \tan x} \over {1 - \tan {\pi \over 3}\tan x}}} \right)} \right)^2}$ $\Rightarrow$ 2y = ${\left( {{{\cot }^{ - 1}}\tan \left( {{\pi \over 3} + x} \right)} \right)^2}$ $\Rightarrow$ 2y = ${\left( {{\pi \over 2} - {{\tan }^{ - 1}}\tan \left( {{\pi \over 3} + x} \right)} \right)^2}$ As x $\in$ $\left( {0,{\pi \over 2}} \right)$ then ${{{\tan }^{ - 1}}\tan \left( {{\pi \over 3} + x} \right)}$ = ${\left( {{\pi \over 3} + x} \right)}$ $\Rightarrow$ 2y = ${\left( {{\pi \over 2} - \left( {{\pi \over 3} + x} \right)} \right)^2}$ $\Rightarrow$ 2y = ${\left( {{\pi \over 6} - x} \right)^2}$ $\therefore$ $2{{dy} \over {dx}} = 2\left( {{\pi \over 6} - x} \right)\left( { - 1} \right)$ $\Rightarrow$ ${{dy} \over {dx}} = \left( {x - {\pi \over 6}} \right)$