化简[(secα-cosα)*(cscα-sinα)]/2sinα*cosα
化简[(secα-cosα)*(cscα-sinα)]/2sinα*cosα
求证:(cosαcscα-sinαsecα)/(cosα+sinα)=cscα-secα
求证(cosαcscα-sinαsecα)/(cosα+sinα)=cscα-secα
化简 [Sin(π+α)\tan(π+α)] ×[ cot(2π-α)\cos(π+α)]×[sec(2π-α)\csc
化简:[cos(α-5π)+2sin(3π-α)]/[csc(3π+α)+sec(5π+α)]
求证:(tanα -cotα )/(secα -cscα )=sinα +cosα
证明 sin^2αtanα+cos^2αcotα+2sinαcosα=secαcscα
证明tan^2α-cot^2α/sin^2α-cos^2α=sec^2α+csc^2α
已知tan^2α+cot^2α+sec^2α+csc^2α=7,则sinαcosα
化简:1/(1+sin²α)+1/(1+cos²α)+1/(1+sec²α)+1/(csc
求证、数学题.求证:tanα-cotα/secα-cscα=sinα-cosα证明题.
证明,[1+sinα)/(1+cosα)]*[(1+secα)/(1+cscα)]=tanα