
如图,在直三棱柱\(ABC-A_{1}B_{1}C_{1}\)中,\(AB⊥AC\),\(AB=AA_{1}=2\),\(AC= \sqrt {2}\),过\(BC\)的中点\(D\)作平面\(ACB_{1}\)的垂线,交平面\(ACC_{1}A_{1}\)于\(E\),则\(BE\)与平面\(ABB_{1}A_{1}\)所成角的正切值为\((\) \()\)
A. \( \dfrac { \sqrt {5}}{5}\)
B. \( \dfrac { \sqrt {5}}{10}\) C. \( \dfrac { \sqrt {10}}{10}\) D. \( \dfrac { \sqrt {10}}{5}\)