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# P2 Trig Proof watch

1. Starting with Sin(A-B) and Cos(A-B)
prove that Tan(A-B) ≡ tanA - tanB/1+tanAtanB
2. (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)
3. (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)
4. got it
5. 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

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