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    Prove that cosecx/cotx + tanx = cosx

    I did 1/sinx (cosx/sinx + sinx/cosx) = cosx/sin^2x + 1/cosx

    Made denominators the same and added, ended up getting 1/sin^2xcosx

    Help please
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    (Original post by G.Y)
    Prove that cosecx/cotx + tanx = cosx

    I did 1/sinx (cosx/sinx + sinx/cosx) = cosx/sin^2x + 1/cosx

    Made denominators the same and added, ended up getting 1/sin^2xcosx

    Help please
    OK, took me a while to notice that you didnt format your original equation correctly and actually meant the second one.

    Anyway, \dfrac{\csc x}{ \cot x + \tan x} = \dfrac{1}{\sin x (\cot x + \tan x)} so you made an error there.
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    (Original post by RDKGames)
    OK, took me a while to notice that you didnt format your original equation correctly and actually meant the second one.

    Anyway, \dfrac{\csc x}{ \cot x + \tan x} = \dfrac{1}{\sin x (\cot x + \tan x)} so you made an error there.
    I don't understand
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    sin = s
    cos = c
    cot = cos/sin
    tan = sin/cos

    (1/s) / [(c/s) + (s/c)] = LHS

    =(1/s) / (c^2/sc + s^2 / sc)
    =(1/s) / [(c^2 + s^2) / sc]
    **c^2 + s^2 = 1**
    **diving by a fraction is the same as multiplying by the reciprocal**
    =1/s * (sc / 1)
    =sc / s
    =c
    = RHS
 
 
 
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