Answer:
[tex]\frac{2\sqrt{3} + 6}{12}[/tex]
Step-by-step explanation:
Simplify the denominator of [tex]\frac{1 + \sqrt{3} }{\sqrt{12} }[/tex] completely.
[tex]=> \frac{1 + \sqrt{3} }{ \sqrt{(4 \times 3)} } = \frac{1 + \sqrt{3} }{\sqrt{4} \times \sqrt{3} } = \frac{1 + \sqrt{3} }{2\sqrt{3} }[/tex]
Now, to rationalise it you need to make the denominator rational.
- Multiply & divide [tex]\frac{1 + \sqrt{3} }{2\sqrt{3} }[/tex] by [tex]2\sqrt{3}[/tex].
[tex]=> \frac{1 + \sqrt{3} }{2\sqrt{3} } \times \frac{2\sqrt{3} }{2\sqrt{3} }[/tex]
[tex]=> \frac{(1 + \sqrt{3})\times 2\sqrt{3} }{2\sqrt{3} \times 2\sqrt{3} }[/tex]
[tex]=> \frac{2\sqrt{3} + 6 }{12}[/tex]
