Deadly Radicals!

$$ $Let~f \left( x \right) =\frac{9^{x}}{9^{x}+3}\text{, then}$

$$ $f \left( 1-x \right) =\frac{9^{1-x}}{9^{1-x}+3}=\frac{\frac{9}{9^{x}}}{\frac{9}{9^{x}}+3}=\frac{9}{9+3 \cdot 9^{x}}=\frac{9}{3 \left( 3+9^{x} \right) }=\frac{3}{9^{x}+3}.$

$$ $\text{Observe that f} \left( x \right) +f \left( 1-x \right) =\frac{9^{x}}{9^{x}+3}+\frac{3}{9^{x}+3}=\frac{9^{x}+3}{9^{x}+3}=1.$

$$ $\text{Now if we define f} \left( x \right) =\sin ^{2} \alpha ~\text{ for some α ϵ R, then the value of }$

$$ $f \left( 1-x \right) =1-f \left( x \right) =1-\sin ^{2} \alpha ~= \cos ^{2} \alpha .$

$$ $\text{For example let }\sin ^{2} \alpha =f \left( \frac{1}{2019} \right) = \frac{9^{\frac{1}{2019}}}{9^{\frac{1}{2019}}+3}\text{, then}$

$$ $\cos ^{2} \alpha =f \left( 1-\frac{1}{2019} \right) =f \left( \frac{2018}{2019} \right) =\frac{9^{\frac{2018}{2019}}}{9^{\frac{2018}{2019}}+3}~.~$

$$ $\text{Now consider the expression}~~\frac{\sqrt[]{1+sin \alpha }+\sqrt[]{1+cos \alpha }}{\sqrt[]{1-sin \alpha }+\sqrt[]{1-cos \alpha }}=$

$$ $= \frac{ \left( \sqrt[]{2}+1 \right) \cos \left( \frac{ \alpha }{2} \right) +sin \left( \frac{ \alpha }{2} \right) }{ \left( \sqrt[]{2}-1 \right) \sin \left( \frac{ \alpha }{2} \right) +cos \left( \frac{ \alpha }{2} \right) } \times \frac{ \left( \sqrt[]{2}+1 \right) }{ \left( \sqrt[]{2}+1 \right) }=\frac{ \left( \sqrt[]{2}+1 \right) \cos \left( \frac{ \alpha }{2} \right) +sin \left( \frac{ \alpha }{2} \right) }{\sin \left( \frac{ \alpha }{2} \right) + \left( \sqrt[]{2}+1 \right) \cos \left( \frac{ \alpha }{2} \right) } \times \left( \sqrt[]{2}+1 \right)$

$$ $=\sqrt[]{2}+1$

$\text{Let g} \left( n \right) =\sqrt[]{f \left( \frac{n}{2019} \right) }=$ $\sqrt[]{\frac{9^{\frac{n}{2019}}}{9^{\frac{n}{2019}}+3}}\text{ ,we conclude that }$

$$ $\frac{\sqrt[]{1+g \left( n \right) }+\sqrt[]{1+g \left( 2019-n \right) }}{\sqrt[]{1-g \left( n \right) }+\sqrt[]{1-g \left( 2019-n \right) }}=\sqrt[]{2}+1.$

$$ $\text{Finally let p} \left( n \right) =\sqrt[]{1+g \left( n \right) }=\sqrt[]{1+\sqrt[]{f \left( \frac{n}{2019} \right) }}=\sqrt[]{1+\sqrt[]{\frac{9^{\frac{n}{2019}}}{9^{\frac{n}{2019}}+3}}}$

$\text{and q} \left( n \right) =\sqrt[]{1-g \left( n \right) }=\sqrt[]{1-\sqrt[]{f \left( \frac{n}{2019} \right) }}=\sqrt[]{1-\sqrt[]{\frac{9^{\frac{n}{2019}}}{9^{\frac{n}{2019}}+3}}}$

$$ $Therefore,~$

$$ $\frac{p \left( 1 \right) +p \left( 2018 \right) }{q \left( 1 \right) +q \left( 2018 \right) }=\frac{p \left( 2 \right) +p \left( 2017 \right) }{q \left( 2 \right) +q \left( 2017 \right) }= \ldots =\frac{p \left( 1009 \right) +p \left( 1010 \right) }{q \left( 1009 \right) +q \left( 1010 \right) }$

$$ $=\frac{ \sum _{1}^{2018}p \left( n \right) }{ \sum _{1}^{2018}q \left( n \right) }=\sqrt[]{2}+1.$

$$ $\text{Hence the value of A is }{(\sqrt[]{2}+1 )}\text{and hence} \left[ A \right] =2.$