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Math Help - [SOLVED] Proving Trig Identities

  1. #1
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    [SOLVED] Proving Trig Identities

    I can't seem to solve this one. Could someone please show me?

    1 - tan^2X
    __________ = cos^2X - sin^2X
    1 + tan^2X

    Thanks!
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  2. #2
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    Quote Originally Posted by demo1 View Post
    I can't seem to solve this one. Could someone please show me?

    1 - tan^2X
    __________ = cos^2X - sin^2X
    1 + tan^2X

    Thanks!
    Multiply the numerator and denominator by \cos^2(x)

    \frac{\cos^2(x)}{\cos^2(x)}\left(\frac{1-\tan^2(x)}{1+\tan^2(x)} \right)=\frac{\cos^2(x)-\sin^2(x)}{\cos^2(x)+\sin^2(x)}=\cos^2(x)-\sin^2(x)
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  3. #3
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    Quote Originally Posted by TheEmptySet View Post
    Multiply the numerator and denominator by \cos^2(x)

    \frac{\cos^2(x)}{\cos^2(x)}\left(\frac{1-\tan^2(x)}{1+\tan^2(x)} \right)=\frac{\cos^2(x)-\sin^2(x)}{\cos^2(x)+\sin^2(x)}=\cos^2(x)-\sin^2(x)

    How do you get from \frac{\cos^2(x)-\sin^2(x)}{\cos^2(x)+\sin^2(x)} to here \cos^2(x)-\sin^2(x)? Thanks btw!
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  4. #4
    Behold, the power of SARDINES!
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    Quote Originally Posted by demo1 View Post
    How do you get from \frac{\cos^2(x)-\sin^2(x)}{\cos^2(x)+\sin^2(x)} to here \cos^2(x)-\sin^2(x)? Thanks btw!
    The pythagorean identity

    \sin^2(x)+\cos^2(x)=1
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  5. #5
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    Quote Originally Posted by TheEmptySet View Post
    The pythagorean identity

    \sin^2(x)+\cos^2(x)=1
    Ah, I really need to review these. Can't believe I didn't see that. Thanks!
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