Answer: Option A
$$\left[ {\,\left( {1 + \frac{1}{{10 + \frac{1}{{10}}}}} \right)\,\left( {1 + \frac{1}{{10 + \frac{1}{{10}}}}} \right)\,\, - \,\,\left( {1 - \frac{1}{{10 + \frac{1}{{10}}}}} \right)\,\left( {1 - \frac{1}{{10 + \frac{1}{{10}}}}} \right)\,} \right]$$              $$ ÷ $$  $$\left[ {\,\left( {1 + \frac{1}{{10 + \frac{1}{{10}}}}} \right)\,\, + \,\,\left( {1 - \frac{1}{{10 + \frac{1}{{10}}}}} \right)\,} \right]$$
$$ {\text{Let }}\left[ {1 + \frac{1}{{10 + \frac{1}{{10}}}}} \right] = a,$$    $${\text{ }}\left[ {1 - \frac{1}{{10 + \frac{1}{{10}}}}} \right] = b $$
$$\eqalign{
& \Rightarrow \left( {{a^2} - {b^2}} \right) \div \left( {a + b} \right) = a - b = ? \cr
& \Rightarrow a = 1 + \frac{{10}}{{101}} \cr
& \Rightarrow a = \frac{{111}}{{101}} \cr
& \Rightarrow b = 1 - \frac{{10}}{{101}} \cr
& \Rightarrow b = \frac{{91}}{{101}} \cr
& \Rightarrow a - b = \frac{{111}}{{101}} - \frac{{91}}{{101}} \cr
& \Rightarrow a - b = \frac{{20}}{{101}} \cr} $$