(Ⅰ)证明:在正方体ABCD-A 1B 1C 1D 1中,点O是BD中点,
∵BC 1=DC 1,BC=DC,
∴C 1O⊥BD,CO⊥BD-------------------(2分)
∵C 1O∩CO=O,C 1O⊂平面C 1OC,CO⊂平面C 1OC,
∴BD⊥平面C 1OC------------------(5分)
∵BD⊂平面BDD 1B 1,∴平面BDD 1B 1⊥平面C 1OC.--------------(7分)
(Ⅱ)由(Ⅰ)可知∠C 1OC是二面角C 1-BD-C的平面角---------------(11分)
则
C 1 C=1,OC=
2
2
∴在Rt△C 1OC中, tan∠ C 1 OC=
C 1 C
OC =
2
故二面角C 1-BD-C的正切值为
2 .---------------(14分)