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    We're using the CGP Textbooks for AQA, and although it gives the answer to this question in the back, it gives quite a lot of leaps and doesn't really explain it self very well.

    Show that (tan(x)+1)(tan(x)-1) is identical to (1/cos^2(x)) - 2

    I get to (1-cos^2(x)/cos^2(x)) -1 and then I'm completely lost as to how you manage to get to
    (1/cos^2(x)) - 2

    I would appreciate any help haha
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    (Original post by CoreyRobertson)
    ...
    \displaystyle \frac{1-\cos^2 x}{\cos^2 x} = \frac{1}{\cos^2 x} - \frac{\cos^2 x}{\cos^2 x} = \frac{1}{\cos^2 x} - 1
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    (Original post by Zacken)
    \displaystyle \frac{1-\cos^2 x}{\cos^2 x} = \frac{1}{\cos^2 x} - \frac{\cos^2 x}{\cos^2 x} = \frac{1}{\cos^2 x} - 1
    That's my problem, I don't get the jump you made int he second step at all... it seems like you just added numbers
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    (Original post by CoreyRobertson)
    That's my problem, I don't get the jump you made int he second step at all... it seems like you just added numbers
    Do you agree that if I have \frac{a + b}{c} this is the same thing as \frac{a}{c} + \frac{b}{c}?
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    (Original post by Zacken)
    Do you agree that if I have \frac{a + b}{c} this is the same thing as \frac{a}{c} + \frac{b}{c}?
    ahhh thanks, I think I see it now, cheers for the help
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    (Original post by CoreyRobertson)
    ahhh thanks, I think I see it now, cheers for the help
    Great! No problem.
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    I would just notice that it is difference of 2 squares. The 2 brackets multiply to give  \displaystyle \tan^2 x -1 = \left (\frac{1}{ \cos^2 x}-1 \right)-1 .
 
 
 
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