1. Simplify Equation- Double Division with Radicals

Hi, I solved tan(5pi/12) using the sum/difference formulas and came up with the following:

(1/sqrt(3) + 1) / (-1/sqrt(3) + 1)

I know it simplifies further down into: 2 + sqrt(3), however I suppose I have forgotten some of the rules of simplification. I don't know how to get it down to that. If someone could explain how to simplify these sorts of expression, that would be much appreciated. Thanks!

2. Use the method of rationalisation, that is multiplying by something such that the denominator becomes a difference of two squares.

$\dfrac{\frac{1}{\sqrt3} + 1}{-\frac{1}{\sqrt3} + 1} \times \dfrac{\frac{1}{\sqrt3} + 1}{\frac{1}{\sqrt3} + 1}$

3. Originally Posted by MrCryptoPrime
Hi, I solved tan(5pi/12) using the sum/difference formulas and came up with the following:

(1/sqrt(3) + 1) / (-1/sqrt(3) + 1)

I know it simplifies further down into: 2 + sqrt(3), however I suppose I have forgotten some of the rules of simplification. I don't know how to get it down to that. If someone could explain how to simplify these sorts of expression, that would be much appreciated. Thanks!
$\dfrac{\frac1{\sqrt{3}}+1}{-\frac1{\sqrt{3}}+1}} = \dfrac{\frac{\sqrt{3} + 1}{\sqrt{3}}}{\frac{\sqrt{3} - 1}{\sqrt{3}}} = \dfrac{\sqrt{3} + 1}{\sqrt{3} - 1}$

$\dfrac{\sqrt{3} + 1}{\sqrt{3} - 1}\cdot \dfrac{\sqrt{3} + 1}{\sqrt{3} + 1} = \dfrac{3+2\sqrt{3} +1}{3-1} = \sqrt{3} + 2$

4. When I multiply do I use FOIL? (e.g. each term is multiplied by all the other terms) I already tried doing what you suggest using FOIL, but it does not seem to help. Could you please elaborate more on the specifics of multiplying that out? Thanks.

5. Nevermind I understand now! Thanks you guys for refreshing my memory!