If $$z=6-2\sqrt3$$, then find the value of $$(\sqrt{z}-\frac{1}{\sqrt{z}})^2$$

Given : $$z=6-2\sqrt3$$ -----------------(i)

To find : $$(\sqrt{z}-\frac{1}{\sqrt{z}})^2$$

= $$(\frac{z-1}{\sqrt{z}})^2=\frac{(z-1)^2}{z}$$ ------------(ii)

From equation (i), => $$z-1=5-2\sqrt3$$

Squaring both sides, we get :

=> $$(z-1)^2=(5-2\sqrt3)^2$$

=> $$(z-1)^2=25+12-20\sqrt3=(37-20\sqrt3)$$ ----------(iii)

Substituting values from equations (i) and (iii) in equation (ii), 

=> $$\frac{(z-1)^2}{z}=\frac{37-20\sqrt3}{6-2\sqrt3}$$

Rationalizing the denominator, we get :

= $$\frac{37-20\sqrt3}{6-2\sqrt3}\times\frac{(6+2\sqrt3)}{(6+2\sqrt3)}$$

= $$\frac{37(6+2\sqrt3)-20\sqrt3(6+2\sqrt3)}{36-12}$$

= $$\frac{222+74\sqrt3-120\sqrt3-120}{24}$$

= $$\frac{102-46\sqrt3}{24}$$

=> Ans - (C)