一道数学题
若sin(180°+α)=1/√10,则[sec(-α)+sin(-α-90°)]/[csc(540°-α)-cos(-α-270°)]=___ 请写出过程,谢谢
sin(180°+α)=-sinα=-1/√10 -180°< α <0 cosα = 3/√10,或-3/√10 cosα = 3sinα,或- 3sinα --[sec(-α)+sin(-α-90°)]/[csc(540°-α)-cos(-α-270°)] =[1/cos(-α)+cos(-α)]/[1/sin(180°-α)-cos(-α+90°)] =[1/cos(α)+cos(α)]/[1/sin(α)-sin(α)] = sina(1+cosa*cosa)/(cosa*cosa*cosa) = sina/cosa * (1+1/(cosa*cosa)) = sina/cosa * (1+1/(cosa*cosa)) = sina/cosa * 19/9 1, =19/3 2, =-19/3
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