已知sin(π-α)-cos(-α)=1/5,求tan[(2k+1)π+α]+cot[(2k+1)π-α](k属於Z)的值

已知sin(π-α)-cos(-α)=1/5,求tan[(2k+1)π+α]+cot[(2k+1)π-α](k属於Z)的值
已知sin(π-α)-cos(-α)=1/5,求tan[(2k+1)π+α]+cot[(2k+1)π-α](k属於Z)的值

已知sin(π-α)-cos(-α)=1/5,求tan[(2k+1)π+α]+cot[(2k+1)π-α](k属於Z)的值
sin(π-α)-cos(-α)=1/5
sinα-cosα=1/5平方
sin²α+cos²α-2sinαcosα=1/25
1-sin2α=1/25
sin2α=24/25
cos2α=±7/25
tan[(2k+1)π+α]+cot[(2k+1)π-α]
=tan[2kπ+π+α]+cot[2kπ+π-α]
=tan(π+α)+cot(π-α)
=tanα-cotα
=sinα/cosα-cosα/sinα
=sin²α/(sinαcosα)-cos²α/(sinαcosα)
=(sin²α-cos²α)/(sinαcosα)
=-cos2α/(1/2sin2α)
=-2cos2α/sin2α
=-2*(±7/25)/(24/25)
=±7/24