Equate it

Square the left hand side of the equation and get $\frac{4}{9} \times \frac{2+6+2\sqrt{12}}{2+\sqrt{3}}$

$= \frac{4}{9} \frac{8+4\sqrt{3}}{2+\sqrt{3}} = \frac{4}{9} \times 4 = (\frac{4}{3})^2$ .

So $\frac{m}{n} = \frac{4}{3}$ and $m+n = 4+3 = \boxed{7}$

Hey.....do you guys know that the same question is in $MTG$ $Maths$ $Olympiad$ $Workbook$ of $Class$ $X$ .

First multiplying with $\sqrt { 2-\sqrt [ ]{ 3 } }$ then multiplying and dividing the expression by $\sqrt { 2 }$ It becomes $\frac { 2 }{ 3 } \frac { (1+\sqrt { 3 } ) }{ 1 } \sqrt { 2-\sqrt { 3 } } \times \sqrt { 2 } =\frac { 2 }{ 3 } \frac { (1+\sqrt { 3 } ) }{ 1 } \sqrt { 4-2\sqrt { 3 } }$ $\sqrt { 4-2\sqrt { 3 } } =\sqrt { 3 } -1$ Thus, the expression becomes 4/3.