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$$\left[ {\frac{{\sqrt 3 + \sqrt 2 }}{{\sqrt 3 - \sqrt 2 }} - \frac{{\sqrt 3 - \sqrt 2 }}{{\sqrt 3 + \sqrt 2 }}} \right]$$ simplifies to = ?
Answer & Solution
Correct Answer:
Option
B
$$\left[ {\frac{{\sqrt 3 + \sqrt 2 }}{{\sqrt 3 - \sqrt 2 }} - \frac{{\sqrt 3 - \sqrt 2 }}{{\sqrt 3 + \sqrt 2 }}} \right]$$
$$ = \frac{{\sqrt 3 + \sqrt 2 }}{{\sqrt 3 - \sqrt 2 }} \times $$ $$\frac{{\sqrt 3 + \sqrt 2 }}{{\sqrt 3 + \sqrt 2 }} - $$ $$\frac{{\sqrt 3 - \sqrt 2 }}{{\sqrt 3 + \sqrt 2 }} \times $$ $$\frac{{\sqrt 3 - \sqrt 2 }}{{\sqrt 3 - \sqrt 2 }}$$
$$\eqalign{ & = \frac{{{{\left( {\sqrt 3 + \sqrt 2 } \right)}^2}}}{{3 - 2}} - \frac{{{{\left( {\sqrt 3 - \sqrt 2 } \right)}^2}}}{{3 - 2}} \cr & = \left( {3 + 2 + 2\sqrt 6 } \right) - \left( {3 + 2 - 2\sqrt 6 } \right) \cr & = 4\sqrt 6 {\text{ }} \cr} $$
$$ = \frac{{\sqrt 3 + \sqrt 2 }}{{\sqrt 3 - \sqrt 2 }} \times $$ $$\frac{{\sqrt 3 + \sqrt 2 }}{{\sqrt 3 + \sqrt 2 }} - $$ $$\frac{{\sqrt 3 - \sqrt 2 }}{{\sqrt 3 + \sqrt 2 }} \times $$ $$\frac{{\sqrt 3 - \sqrt 2 }}{{\sqrt 3 - \sqrt 2 }}$$
$$\eqalign{ & = \frac{{{{\left( {\sqrt 3 + \sqrt 2 } \right)}^2}}}{{3 - 2}} - \frac{{{{\left( {\sqrt 3 - \sqrt 2 } \right)}^2}}}{{3 - 2}} \cr & = \left( {3 + 2 + 2\sqrt 6 } \right) - \left( {3 + 2 - 2\sqrt 6 } \right) \cr & = 4\sqrt 6 {\text{ }} \cr} $$
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