Question

Hello everybody ✌
Pls rationalize the above sum..

Answer 1

$<b>$Answer :

$= > \frac{2 + \sqrt{3} }{2 - \sqrt{3} } + \frac{2 - \sqrt{3} }{2 + \sqrt{3} } \\ \\ = > \frac{2 + \sqrt{3} }{2 - \sqrt{3} } \times \frac{2 + \sqrt{3} }{2 + \sqrt{3} } + \frac{2 - \sqrt{3} }{2 + \sqrt{3} } \times \frac{2 - \sqrt{3} }{2 - \sqrt{3} } \\ \\ = > \frac{ {(2 + \sqrt{3} )}^{2} }{ {(2)}^{2} - {( \sqrt{3} )}^{2} } + \frac{ {(2 - \sqrt{3} )}^{2} }{ {(2)}^{2} - {( \sqrt{3}) }^{2} } \\ \\ = > \frac{ {(2)}^{2} + {( \sqrt{3} )}^{2} + 2 \times 2 \times \sqrt{3} }{4 - 3} + \frac{ {(2)}^{2} + {( \sqrt{3}) }^{2} - 2 \times 2 \times \sqrt{3} }{4 - 3} \\ \\ = > \frac{4 + 3 + 4 \sqrt{3} }{1} + \frac{4 + 3 - 4 \sqrt{3} }{1} \\ \\ = > \frac{7 + 4 \sqrt{3} }{1} + \frac{7 - 4 \sqrt{3} }{1} \\ \\ = > (7 + 4 \sqrt{3} ) + (7 - 4 \sqrt{3} ) \\ \\ = > 7 + 4 \sqrt{3} + 7 - 4 \sqrt{3} \\ \\ = > 7 + 7 \\ \\ = > 14$

$\color{pink}\underline\textbf{Hence, 14 is your required answer. }$

Tysm for the question!

Answer 2

Here, I used a shortcut , when the denominators have same numbers with different signs you can do like this ( by cross multiply ).

》》 First attachment is the solution.

》》 Second attachment is the simplification and Identity used in the solution.

》》》 The other answer given by Beautiful5665 is of correct method without shortcuts.