若sin[α-(2n+1)π/2]=3/5,α∈(0,π/2)∪(π/2,π),则tanα+1/tanα=?若sin[α-(2n+1)π/2]=3/5,α∈(0,π/2)∪(π/2,π),则tanα+1/tanα=

若sin[α-(2n+1)π/2]=3/5,α∈(0,π/2)∪(π/2,π),则tanα+1/tanα=?若sin[α-(2n+1)π/2]=3/5,α∈(0,π/2)∪(π/2,π),则tanα+1/tanα=
若sin[α-(2n+1)π/2]=3/5,α∈(0,π/2)∪(π/2,π),则tanα+1/tanα=?
若sin[α-(2n+1)π/2]=3/5,α∈(0,π/2)∪(π/2,π),则tanα+1/tanα=

若sin[α-(2n+1)π/2]=3/5,α∈(0,π/2)∪(π/2,π),则tanα+1/tanα=?若sin[α-(2n+1)π/2]=3/5,α∈(0,π/2)∪(π/2,π),则tanα+1/tanα=
若n为奇数,则
sin[x-(2n+1)pai/2]=sin(x+pai/2)=cosx=3/5;xpai/2;tanx=-4/3；
此时答案为-25/12

sin[x-(2n+1)pai/2]=sin(x+pai/2)=cosx=3/5;x若n为偶数，则
sin[x-(2n+1)pai/2]=sin(x-pai/2)=-cosx=3/5;x>pai/2;tanx=-4/3；