The simplified value of :

$\left\{\begin{array}{l}\left(1+\frac{1}{10+\frac{1}{10}}\right)\left(1+\frac{1}{10+\frac{1}{10}}\right) \\ \left(1-\frac{1}{10+\frac{1}{10}}\right)\left(1-\frac{1}{10+\frac{1}{10}}\right)\end{array}\right\} \div\left\{\left(\begin{array}{l}\left.\left.1+\frac{1}{10+\frac{1}{10}}\right)\left(1-\frac{1}{10+\frac{1}{10}}\right)\right\}\end{array}\right\}\right.$

Let, $a=1+\frac{1}{10+\frac{1}{10}}=1+\frac{1}{\frac{100+1}{10}}=1+\frac{10}{101}$

$=\frac{101+10}{101}=\frac{111}{101}$

Again,

$b=1-\frac{1}{10+\frac{1}{10}}=1-\frac{1}{\frac{100+1}{10}}$

$=1-\frac{10}{101}=\frac{101-10}{101}=\frac{91}{101}$

$\therefore$ Expression $=\left(a^{2}-b^{2}\right) \div a b$

$=\{(a+b)(a-b)\} \div a b$

$=\left(\frac{111}{101}+\frac{91}{101}\right)\left(\frac{111}{101}-\frac{91}{101}\right) \div\left(\frac{111}{101} \times \frac{91}{101}\right)$

$=\frac{202}{101} \times \frac{20}{101} \times \frac{101 \times 101}{111 \times 91}=\frac{4040}{10101}$