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    prove cosec(theta) - sin(theta) = cot(theta)cos(theta)

    i chose to start with LHS

    cosec(theta) - sin(theta) = 1/sin(theta) - sin(theta)

    = (1 - sin^2(theta)) / sin(theta) = (cos^2(theta)) / sin(theta)

    I'm guessing that this is correct, but i don't understand how the cos(theta) can be isolated from cot(theta) to get it = to RHS. Can someone explain?
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    Use the identity cot(theta) = cos(theta)/sin(theta):

    cos^2(theta) / sin(theta)

    = cos(theta)/sin(theta) * cos(theta)

    = cot(theta)cos(theta)
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    (Original post by AndyOC)
    prove cosec(theta) - sin(theta) = cot(theta)cos(theta)

    i chose to start with LHS

    cosec(theta) - sin(theta) = 1/sin(theta) - sin(theta)

    = (1 - sin^2(theta)) / sin(theta) = (cos^2(theta)) / sin(theta)

    I'm guessing that this is correct, but i don't understand how the cos(theta) can be isolated from cot(theta) to get it = to RHS. Can someone explain?
    Remember that

    \tan x = \dfrac{\sin x}{\cos x}

    and so

    \displaystyle \cot x = \dfrac{1}{\tan x} = \dfrac{\cos x }{\sin x}
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    (Original post by Indeterminate)
    Remember that

    \tan x = \dfrac{\sin x}{\cos x}

    and so

    \displaystyle \cot x = \dfrac{1}{\tan x} = \dfrac{\cos x }{\sin x}
    (Original post by Pronged Lily)
    Use the identity cot(theta) = cos(theta)/sin(theta):

    cos^2(theta) / sin(theta)

    = cos(theta)/sin(theta) * cos(theta)

    = cot(theta)cos(theta)
    Ah yeah of course, thanks for the help!
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    1/sin(theta) - sin(theta)

    = (1 - sin^2(theta)) / sin(theta)


    How did you go from this step to this as if you times both sides by Sin(Theta) it will be only 1-sin(squared)(Theta).
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    (Original post by Chocolatemousse)
    1/sin(theta) - sin(theta)

    = (1 - sin^2(theta)) / sin(theta)


    How did you go from this step to this as if you times both sides by Sin(Theta) it will be only 1-sin(squared)(Theta).
    It's not an equation where you times both sides by something, it's just combining fractions:

     1/\sin \theta - \sin \theta = 1/\sin \theta - \sin^2 \theta / \sin \theta
 
 
 
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