simplify this expression 2-√(2+√3)/√(2+√3)

Question

2-√(2+√3)/√(2+√3) I need to simplify this. Can I multiplay with 2-√3 the numerator and dominator? I need your help.

Answer 1

$$\frac{2-\sqrt{2+\sqrt3}}{\sqrt{2+\sqrt3}}=\frac2{\sqrt{2+\sqrt3}}-1=\frac{2\cdot\color{red}{\sqrt{2-\sqrt3}}}{\sqrt{2+\sqrt3}\cdot\color{red}{\sqrt{2-\sqrt3}}}-1=2\cdot\color{red}{\sqrt{2-\sqrt3}}-1.$$ Likewise, for every $a\geqslant1$ and every $b$, $$\frac{b-\sqrt{a+\sqrt{a^2-1}}}{\sqrt{a+\sqrt{a^2-1}}}=b\cdot\sqrt{a-\sqrt{a^2-1}}-1.$$

Answer 2

$$\frac{2-\sqrt{2+\sqrt3}}{\sqrt{2+\sqrt3}}=\frac{2-\sqrt{2+\sqrt3}}{\sqrt{2+\sqrt3}}\frac{\sqrt{2-\sqrt3}}{\sqrt{2-\sqrt3}}=\frac{2\sqrt{2-\sqrt3}-\sqrt{2+\sqrt3}\sqrt{2-\sqrt3}}{\sqrt{2+\sqrt3}\sqrt{2-\sqrt3}}=$$ $$=\frac{2\sqrt{2-\sqrt3}-\sqrt{(2+\sqrt3)(2-\sqrt3)}}{\sqrt{(2+\sqrt3)(2-\sqrt3)}}=\frac{2\sqrt{2-\sqrt3}-\sqrt{2^2-(\sqrt3)^2}}{\sqrt{2^2-(\sqrt3)^2}}=$$ $$=\frac{2\sqrt{2-\sqrt3}-\sqrt{4-3}}{\sqrt{4-3}}=\frac{2\sqrt{2-\sqrt3}-\sqrt{1}}{\sqrt{1}}=2\sqrt{2-\sqrt3}-1$$

Answer 3

If your expression is $\frac{2-\sqrt{2+\sqrt{3}}}{\sqrt{2+\sqrt{3}}}$, it must be $\sqrt{2-\sqrt{3}}$.