作业帮若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=
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![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=](/uploads/image/z/10229201-17-1.jpg?t=%E8%8B%A5sin%5B%CE%B1-%282n%2B1%29%CF%80%2F2%5D%3D3%2F5%2C%CE%B1%E2%88%88%280%2C%CF%80%2F2%29%E2%88%AA%28%CF%80%2F2%2C%CF%80%29%2C%E5%88%99tan%CE%B1%2B1%2Ftan%CE%B1%3D%3F%E8%8B%A5sin%5B%CE%B1-%282n%2B1%29%CF%80%2F2%5D%3D3%2F5%2C%CE%B1%E2%88%88%280%2C%CF%80%2F2%29%E2%88%AA%28%CF%80%2F2%2C%CF%80%29%2C%E5%88%99tan%CE%B1%2B1%2Ftan%CE%B1%3D)
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
sin[x-(2n+1)pai/2]=sin(x-pai/2)=-cosx=3/5;x>pai/2;tanx=-4/3；