参考例题：
在正方体AC1中 E.F分别是BC,CC1是中点求二面角F-DE-C的正切值[qid:384641977][ QID:384641977]
设二面角F-DE-C=a 正方体AC1的边长为m
S(DEC)=1/4 m^2
DE=DF=√5m/2 EF=√2m/2 p=1/2(DE+DF+EF) = (√5/2+√2/4)m
S(DEF)= √[p(p-DE)(p-DF)(p-EF)] = 3/4 m^2
cosa=S(DEC)/S(DEF) =1/3
sina = √（1-1/9）=2√2/3
tana=sina/cosa = 2√2