<s>[INST] Show that $\sqrt{2-\sqrt{3}}=\frac{1}{2}(\sqrt{6}-\sqrt{2})$ 
Prove the following:
  $\sqrt{2-\sqrt{3}}=\frac{1}{2}(\sqrt{6}-\sqrt{2})$

How do you go about proving this?
 [/INST] \begin{align}\sqrt{2-\sqrt{3}} &=\frac{\sqrt{2}\sqrt{2-\sqrt{3}}}{\sqrt{2}}\\
 &=\frac {\sqrt { 4-2\sqrt{3}}}{\sqrt{2}} \\
&=\frac {\sqrt{1-2\sqrt{3} +{ \left( \sqrt{3}\right)}^{2} }  }{ \sqrt {2}  } \\
&=\frac{\sqrt { { \left( 1-\sqrt{3}\right)}^{2}}}{ \sqrt{2}}\\
 &=\frac{\sqrt {3} -1 }{ \sqrt {2}  }\\
& = \frac{\sqrt{6}-\sqrt{2}}{2} \end{align}
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