If sin A - A = 3 - 12 , then the value of sin A. cosA is?

Answer: Option D

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