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    Prove that 1/cosA - cosA = sinAtanA
    1/(sinA/tanA) - sinA/tanA
    tanA/sinA - sinA/tanA
    TanA/sinAtanA - sinA/sinAtanA
    sinA - tanA

    Where have i gone wrong?
    sorry if this is hard to read
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    (Original post by physicshelpme)
    Prove that 1/cosA - cosA = sinAtanA
    1/(sinA/tanA) - sinA/tanA
    tanA/sinA - sinA/tanA
    TanA/sinAtanA - sinA/sinAtanA
    sinA - tanA

    Where have i gone wrong?
    sorry if this is hard to read
    Just get a common denominator on the LHS.
    That will leave you 1-cos^2(A)/cos(A)

    What does that now give?.......
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    (Original post by physicshelpme)
    Prove that 1/cosA - cosA = sinAtanA
    1/(sinA/tanA) - sinA/tanA
    tanA/sinA - sinA/tanA
    TanA/sinAtanA - sinA/sinAtanA
    sinA - tanA

    Where have i gone wrong?
    sorry if this is hard to read
    You should have combined the LHS, and then proceeded to show that it equals the RHS.

    \dfrac{a}{b} - b = \dfrac{a-b^2}{b}
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    Also, in your third line, you can't just randomly multiply everything by 1/tan(A) and 1/sin(A). Whatever you do to the bottom, you have to do to the top.
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    (Original post by Indeterminate)
    You should have combined the LHS, and then proceeded to show that it equals the RHS.

    \dfrac{a}{b} - b = \dfrac{a-b^2}{b}
    why does it become b^2?
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    (Original post by physicshelpme)
    why does it become b^2?
    actually i understand, thank you
 
 
 
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