Simplify : $$\left[ {\left( {1 + \frac{1}{{10 + \frac{1}{{10}}}}} \right) \times \left( {1 + \frac{1}{{10 + \frac{1}{{10}}}}} \right) - \left( {1 - \frac{1}{{10 + \frac{1}{{10}}}}} \right) \times \left( {1 - \frac{1}{{10 + \frac{1}{{10}}}}} \right)} \right] \div \left[ {\left( {1 + \frac{1}{{10 + \frac{1}{{10}}}}} \right) + \left( {1 - \frac{1}{{10 + \frac{1}{{10}}}}} \right)} \right] = ?$$

Solution (By Examveda Team)

$$\eqalign{ & {\text{According to question,}} \cr & \frac{{\,\left[ {\left( {1 + \frac{1}{{10 + \frac{1}{{10}}}}} \right) \times \left( {1 + \frac{1}{{10 + \frac{1}{{10}}}}} \right) - \left( {1 - \frac{1}{{10 + \frac{1}{{10}}}}} \right) \times \left( {1 - \frac{1}{{10 + \frac{1}{{10}}}}} \right)} \right]}}{{\,\left[ {\left( {1 + \frac{1}{{10 + \frac{1}{{10}}}}} \right) + \left( {1 - \frac{1}{{10 + \frac{1}{{10}}}}} \right)} \right]}} \cr & {\text{let ,}}1 + \frac{1}{{10 + \frac{1}{{10}}}} = \frac{{111}}{{101}} = a \cr & \,\,\,\,\,\,\,\,1 - \frac{1}{{10 + \frac{1}{{10}}}} = \frac{{91}}{{101}} = b \cr & \Rightarrow \frac{{{a^2} - {b^2}}}{{a + b}}\left[ {\because {a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)} \right] \cr & \Rightarrow \frac{{\left( {a + b} \right)\left( {a - b} \right)}}{{a + b}} \cr & \Rightarrow \left( {a - b} \right) \cr & \Rightarrow \frac{{111}}{{101}} - \frac{{91}}{{101}} \cr & \Rightarrow \frac{{20}}{{101}} \cr} $$