If $${\text{sin A}} - \cos {\text{A}}$$   = $$\frac{{\sqrt 3 - 1}}{2}{\text{,}}$$   then the value of $${\text{sin A}}.{\text{cosA}}$$   is?

$$\eqalign{ & {\text{sin A}} - \cos {\text{A}} = \frac{{\sqrt 3 - 1}}{2} \cr & {\bf{Shortcut \,\,method:}} \cr & {\text{Put, }}\theta = {60^ \circ } \cr & \Rightarrow {\text{sin A}} - \cos {\text{A}} = \frac{{\sqrt 3 - 1}}{2} \cr & \Rightarrow {\text{sin }}{60^ \circ } - \cos {60^ \circ } = \frac{{\sqrt 3 - 1}}{2} \cr & \Rightarrow \frac{{\sqrt 3 }}{2} - \frac{1}{2} = \frac{{\sqrt 3 - 1}}{2} \cr & \Rightarrow \frac{{\sqrt 3 - 1}}{2} = \frac{{\sqrt 3 - 1}}{2}({\text{Matched}}) \cr & Hence, \cr & {\text{sin A}}.cos{\text{A}} \cr & \Rightarrow \frac{{\sqrt 3 }}{2} \times \frac{1}{2} \cr & \Rightarrow \frac{{\sqrt 3 }}{4} \cr & \cr & {\bf{Alternate:}} \cr & {\text{sin A}} - \cos {\text{A}} = \frac{{\sqrt 3 - 1}}{2} \cr & {\text{Squaring both side,}} \cr & \Rightarrow {\text{ si}}{{\text{n}}^2}{\text{ A + }}{\cos ^2}{\text{A}} - {\text{2}}{\text{.sin A}}.cos{\text{A}} = {\left( {\frac{{\sqrt 3 - 1}}{2}} \right)^2} \cr & \Rightarrow 1 - 2{\text{sin A}}.cos{\text{A = }}\frac{{3 + 1 - 2\sqrt 3 }}{4} \cr & \Rightarrow 2{\text{sin A}}.cos{\text{A}} = 1 - 2\frac{{\left( {2 - \sqrt 3 } \right)}}{4} \cr & \Rightarrow 2{\text{sin A}}.cos{\text{A}} = \frac{{2 - 2 + \sqrt 3 }}{2} \cr & \Rightarrow {\text{sin A}}.cos{\text{A}} = \frac{{\sqrt 3 }}{4} \cr} $$