已知 sin(A+B)=1\2,sin(A-B)=1\3,求tan(A+B)-tanA-tanB\tanB*tanB*t

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  • sin(a+b)=sinacosb+cosasinb=1/2……(1)

    sin(a-b)=sinacosb-cosasinb=1/3……(2)

    (1)+(2)

    2sinacosb=5/6,

    sinacosb=5/12

    (1)-(2)

    2cosasinb=1/6,

    cosasinb=1/12

    tan(a+b)-tana-tanb

    =tan(a+b)-(tana+tanb)

    =tan(a+b)-[tan(a+b)*(1-tanatanb]

    =tana*tanb*tan(a+b)

    (tan(a+b)-tana-tanb)/(tan^2b*tan(a+b))

    =tana*tanb*tan(a+b)/(tan^2b*tan(a+b))

    =tana*tanb/tan^2b

    =tana/tanb

    =sinacosb/cosasinb

    =(5/12)/(1/12)

    =5