$$\left( {\frac{{2 + \sqrt 3 }}{{2 - \sqrt 3 }} + \frac{{2 - \sqrt 3 }}{{2 + \sqrt 3 }} + \frac{{\sqrt 3 - 1}}{{\sqrt 3 + 1}}} \right)$$ simplifies to = ?
A. $$16 - \sqrt 3 $$
B. $$4 - \sqrt 3 $$
C. $$2 - \sqrt 3 $$
D. $$2 + \sqrt 3 $$
Answer: Option A
Solution (By Examveda Team)
Given expression,$$ = \frac{{\left( {2 + \sqrt 3 } \right)}}{{\left( {2 - \sqrt 3 } \right)}} \times \frac{{\left( {2 + \sqrt 3 } \right)}}{{\left( {2 + \sqrt 3 } \right)}}$$ $$ + \frac{{\left( {2 - \sqrt 3 } \right)}}{{\left( {2 + \sqrt 3 } \right)}}$$ $$ \times \frac{{\left( {2 - \sqrt 3 } \right)}}{{\left( {2 - \sqrt 3 } \right)}}$$ $$ + \frac{{\left( {\sqrt 3 - 1} \right)}}{{\left( {\sqrt 3 + 1} \right)}}$$ $$ \times \frac{{\left( {\sqrt 3 - 1} \right)}}{{\left( {\sqrt 3 - 1} \right)}}$$
$$ = \frac{{{{\left( {2 + \sqrt 3 } \right)}^2}}}{{\left( {4 - 3} \right)}} + \frac{{{{\left( {2 - \sqrt 3 } \right)}^2}}}{{\left( {4 - 3} \right)}}$$ $$ + \frac{{{{\left( {\sqrt 3 - 1} \right)}^2}}}{{\left( {3 - 1} \right)}}$$
$$ = \left[ {{{\left( {2 + \sqrt 3 } \right)}^2} + {{\left( {2 - \sqrt 3 } \right)}^2}} \right]$$ $$ + \frac{{4 - 2\sqrt 3 }}{2}$$
$$ = 2\left( {4 + 3} \right) + 2 - \sqrt 3 $$
$$ = 16 - \sqrt 3 $$
This Question Belongs to Arithmetic Ability >> Square Root And Cube Root
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