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anonymous

  • 5 years ago

prove using identities that ((sec^2)x)/(((sec^2)x)-1)=(csc^2)x

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  1. bahrom7893
    • 5 years ago
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    Okay, so let me try this: did you mean: \[Sec^2(x)/(Sec^2x - 1) = Csc^2(x)\]

  2. anonymous
    • 5 years ago
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    yea. sorry i don't know how get it like that on my computer

  3. bahrom7893
    • 5 years ago
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    okay so now workin on it

  4. anonymous
    • 5 years ago
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    thank you so much

  5. bahrom7893
    • 5 years ago
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    There are a few Sec and Csc identities, but I never remember them so im just gonna convert em to Sin and Cos: Sec(x) = 1/Cos(x) Csc(x) = 1/Sin(x)

  6. bahrom7893
    • 5 years ago
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    \[[1/Cos^2x] / [ (1/Cos^2x)-1 ] = 1/Sin^2x\]

  7. bahrom7893
    • 5 years ago
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    Now let's simplify this part: \[(1/Cos^2x) - 1\] = \[(1-Cos^2x)/Cos^2x\]

  8. anonymous
    • 5 years ago
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    wait. how did you do that?

  9. bahrom7893
    • 5 years ago
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    find the common denominator

  10. bahrom7893
    • 5 years ago
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    1 = Cos^2 / Cos^2... get it?

  11. anonymous
    • 5 years ago
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    oh . yea, sorry. i got it

  12. bahrom7893
    • 5 years ago
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    k

  13. bahrom7893
    • 5 years ago
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    So now you have to divide 1/Cos^2x by (1-Cos^2x)/Cos^2x which is the same as multiplying by its inverse

  14. bahrom7893
    • 5 years ago
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    \[[1/Cos^2x] * [Cos^2x/(1-Cos^2x)] = 1/Sin^2x\]

  15. bahrom7893
    • 5 years ago
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    Cancel out cos^2s

  16. bahrom7893
    • 5 years ago
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    \[1/(1-Cos^2x) = 1/Sin^2x\]

  17. bahrom7893
    • 5 years ago
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    Cross multiply:

  18. anonymous
    • 5 years ago
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    wait. can't you just say 1 / (sin^2x) =csc^2x

  19. anonymous
    • 5 years ago
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    and then be done? or is that skipping a step?

  20. bahrom7893
    • 5 years ago
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    yeah, but i prefer keeping everything in sin and cosine

  21. bahrom7893
    • 5 years ago
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    bcuz its easy to simplify

  22. bahrom7893
    • 5 years ago
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    \[Sin^2x = 1 - Cos^2x\]

  23. bahrom7893
    • 5 years ago
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    See that's proved. Sin^2x + Cos^2x = 1!!!!

  24. bahrom7893
    • 5 years ago
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    get it?

  25. anonymous
    • 5 years ago
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    yes. thank you so much!

  26. bahrom7893
    • 5 years ago
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    k off to helpin someone else.. any questions: email me bahrom.cfo@gmail.com and if u liked my explanation, bcum a fan YAY!

  27. anonymous
    • 5 years ago
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    deff will. that was so helpful. thanks again!

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