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    Starting with Sin(A-B) and Cos(A-B)
    prove that Tan(A-B) ≡ tanA - tanB/1+tanAtanB
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    (Original post by ThugzMansion7)
    Starting with Sin(A-B) and Cos(A-B)
    prove that Tan(A-B) ≡ tanA - tanB/1+tanAtanB
    tan(A-B)=sin(A-B)/cos(A-B)=[sinAcosB-cosAsinB]/[cosAcosB+sinAsinB]=[tanA-tanB]/[1+tanAtanB] (dividing top and bottom by cosAcosB)
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    (Original post by Gaz031)
    tan(A-B)=sin(A-B)/cos(A-B)=[sinAcosB-cosAsinB]/[cosAcosB+sinAsinB]=[tanA-tanB]/[1+tanAtanB] (dividing top and bottom by cosAcosB)
    :confused:
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    got it
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    If Sin(A-B) = SinACosB-CosASinB
    and Cos(A-B) = CosACosB+SinASinB

    So Tan(A-B) = (SinACosB-CosASinB)/(CosACosB+SinASinB)


    Tan(A-B) = (TanA - TanB)/(1-TanATanB) by ÷by cosAcosB


    Just abit more spread out
 
 
 
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