x^2-(\tan \alpha+\cot \alpha)x+1 = 0

x=\dfrac{\tan \alpha+\cot \alpha\pm\sqrt{(\tan \alpha+\cot \alpha)^2-4} }{2}

Since one of roots is

2+\sqrt{3} =\dfrac{4+\sqrt{12} }{2}

Cancel the one witt negative radical

\tan \alpha+\cot \alpha=4

(\tan \alpha+\cot \alpha)^2-4=12

\dfrac{\sin \alpha }{\cos \alpha } +\dfrac{\cos \alpha }{\sin \alpha } =4

(\dfrac{\sin \alpha }{\cos \alpha } +\dfrac{\cos \alpha }{\sin \alpha } )^2=16

Therefore \sin \alpha \cdotp \cos \alpha=\dfrac{1}{4}