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kshoaf

  • 2 years ago

how does (sec^2(x)-1)/(sec^2(x))=1-cos^2(x)

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  1. Hashir
    • 2 years ago
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    |dw:1375066992129:dw|

  2. Hashir
    • 2 years ago
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    |dw:1375067024532:dw|

  3. Hashir
    • 2 years ago
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    |dw:1375067051763:dw|

  4. kshoaf
    • 2 years ago
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    lol thanks im thinking to hard

  5. flixoe
    • 2 years ago
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    First, split the left-side fraction into two: (sec^2(x)-1)/(sec^2(x))=1-cos^2(x) (sec^2(x)/(sec^2(x) - 1/(sec^2(x)) = 1 - cos^2(x) Then, simplify the first item to 1: (sec^2(x)/(sec^2(x) - 1/(sec^2(x)) = 1 - cos^2(x) 1 - 1/(sec^2(x)) = 1 - cos^2(x) Lastly, simplify sec^2(x) into cosine: 1 - 1/(sec^2(x)) = 1 - cos^2(x) 1 - cos^2(x) = 1 - cos^2(x) And it's equal.

  6. Hashir
    • 2 years ago
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    @kshoaf : LOL... pleasure :)

  7. Hashir
    • 2 years ago
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    thinking hard is not to good for health ;)

  8. kshoaf
    • 2 years ago
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    not at all...especially with a final tomorrow. How am I supposed to do the hard stuff when I'm over thinking the simple steps

  9. Hashir
    • 2 years ago
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    aw... you will ace it :) dont worry

  10. Hashir
    • 2 years ago
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    the best thing to do before a maths test is relax :D

  11. kshoaf
    • 2 years ago
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    Thanks!

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