If $$a\left( {2 + \sqrt 3 } \right)$$   = $$b\left( {2 - \sqrt 3 } \right)$$   = 1, then the value of $$\frac{1}{{{a^2} + 1}}$$  + $$\frac{1}{{{b^2} + 1}}$$  = ?

Solution (By Examveda Team)

$$\eqalign{ & a\left( {2 + \sqrt 3 } \right) = b\left( {2 - \sqrt 3 } \right) = 1 \cr & a = \frac{1}{{\left( {2 + \sqrt 3 } \right)}} \cr & b = \frac{1}{{\left( {2 - \sqrt 3 } \right)}} \cr & \Rightarrow a = \frac{1}{b} \cr & \Rightarrow \frac{1}{{{a^2} + 1}} + \frac{1}{{{b^2} + 1}} \cr & \Rightarrow \frac{1}{{\frac{1}{{{b^2}}} + 1}} + \frac{1}{{{b^2} + 1}} \cr & \Rightarrow \frac{1}{{\frac{{1 + {b^2}}}{{{b^2}}}}} + \frac{1}{{{b^2} + 1}} \cr & \Rightarrow \frac{{{b^2}}}{{{b^2} + 1}} + \frac{1}{{{b^2} + 1}} \cr & \Rightarrow \frac{{{b^2} + 1}}{{{b^2} + 1}} \cr & \Rightarrow 1 \cr} $$