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    Prove the identity \tan 2x - \tan x = \tan x \sec 2x please help me im soo confused .
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    What have you tried?
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    ive tried a few things but not really making much progress
    changed the  sec2x into  1 + tan2x

    and on the left hand side would you change the tan2x into  \frac {2tan A} {1 - tan^2A}
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    Don't rush off with double-angle formulae (it turns out you don't even need them here).

    First convert the LHS and RHS so that they're in terms of sine and cosine. It should look like this:

    \dfrac{\sin 2x}{\cos 2x} - \dfrac{\sin x}{\cos x} = \dfrac{\sin x}{\cos x} \dfrac{1}{\cos 2x}

    I've not done anything here; this is just a different way of writing it. You want to go from the LHS to the RHS [you could go the other way round but that's not as easy].

    Now start by combining the two fractions on the LHS into one fraction (by putting them over a common denominator). When you do this, you might notice something familiar.

    If you don't notice anything
    Write out the expansion of \sin (A-B) and compare it to the numerator of your fraction.
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    You are getting muddled between \sec^2 x and \sec 2x. They are not the same.
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    oh yeah thanks!
 
 
 
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Updated: September 29, 2010
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