How to simplify the nested radical $\sqrt{1 - \frac{\sqrt{3}}{2}}$ by hand?

One may write $$ \begin{align} \sqrt{1 - \dfrac{\sqrt{3}}{2}}&=\dfrac1{\sqrt{2}}\sqrt{2 - \sqrt{3}}\\\\ &=\dfrac{\sqrt{2}}2\sqrt{2 - \sqrt{3}}\\\\ &=\dfrac12\sqrt{4 - 2\sqrt{3}}\\\\ &=\dfrac12\sqrt{(\sqrt{3}-1)^2}\\\\ &=\dfrac{\sqrt{3}-1}2. \end{align} $$

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

Proposed another way to collection

You can also use formulas $$\sqrt{a\pm\sqrt b}=\sqrt{\frac{a + \sqrt{a^2-b}}{2}}\pm\sqrt{\frac{a - \sqrt{a^2-b}}{2}}$$ Then $$\sqrt{1-\frac{\sqrt3}{2}}=\frac12\sqrt{4-\sqrt{12}}=\frac12 \cdot\left(\sqrt{\frac{4 + \sqrt{4^2-12}}{2}}-\sqrt{\frac{4 - \sqrt{4^2-12}}{2}} \right)=$$ $$=\frac12 \left(\sqrt{\frac{4 + \sqrt{4}}{2}}-\sqrt{\frac{4 - \sqrt{4}}{2}} \right)=\frac12 \left(\sqrt{\frac{6}{2}}-\sqrt{\frac{2}{2}} \right)=\frac 12(\sqrt3-1)$$