In $$\triangle$$ABC, right angled at B, if cot A = $$\frac{1}{2}$$, then the value of $$\frac{\sin A(\cos C + \cos A)}{\cos C(\sin C - \sin A)}$$ is

cot A = $$\frac{\text{Adjacent side}}{\text{Opposite side}}$$ $$\frac{1}{2}$$

$$\frac{\sin A(\cos C + \cos A)}{\cos C(\sin C - \sin A)}$$ = $$\frac{\frac{2}{\sqrt{5}}\left(\frac{2}{\sqrt{5}}+\frac{1}{\sqrt{5}}\right)}{\frac{2}{\sqrt{5}}\left(\frac{1}{\sqrt{5}}-\frac{2}{\sqrt{5}}\right)}$$

= $$\frac{\left(\frac{3}{\sqrt{5}}\right)}{\left(-\frac{1}{\sqrt{5}}\right)}$$

= -3

Hence, the correct answer is Option B