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    Hi guys, I was a little stuck on this trig proof. I was wondering if anyone could help me:
    (cotx-cosx)/(1-sinx) = cotx

    Thanks
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    (Original post by loooolo12345)
    Hi guys, I was a little stuck on this trig proof. I was wondering if anyone could help me:
    (cotx-cosx)/(1-sinx) = cotx

    Thanks
    Try converting everything to sines and cosines - what do you get? Post your working so we can help.
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    ok thanks

    So: [(cosx/sinx)-cosx]/(1-sinx)
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    (Original post by loooolo12345)
    ok thanks

    So: [(cosx/sinx)-cosx]/(1-sinx)
    Common denominators on the numerator:

    \displaystyle \frac{\frac{\cos x}{\sin x} - \cos x}{1-\sin x} = \frac{\frac{\cos x - \cos x \sin x}{\sin x}}{1-\sin x} = \frac{\frac{\cos x(1-\sin x)}{\sin x}}{1- \sin x} = \cdots
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    (Original post by Zacken)
    Common denominators on the numerator:

    \displaystyle \frac{\frac{\cos x}{\sin x} - \cos x}{1-\sin x} = \frac{\frac{\cos x - \cos x \sin x}{\sin x}}{1-\sin x} = \frac{\frac{\cos x(1-\sin x)}{\sin x}}{1- \sin x} = \cdots
    Ah ok so can you do:

    [cotx(1-sinx)]/(1-sinx)

    then cancel out 1-sinx?
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    (Original post by loooolo12345)
    Ah ok so can you do:

    [cotx(1-sinx)]/(1-sinx)

    then cancel out 1-sinx?
    Perfecto.
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    (Original post by Zacken)
    Perfecto.
    Thats great! thank you so much for your help!
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    (Original post by loooolo12345)
    Thats great! thank you so much for your help!
    No problem. :-)
 
 
 
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