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zhenga2 Group Title

cscθ-cotθ=sinθ/(1+cosθ) im working on the left side and i got: (1/sinθ) - cosθ/sinθ = (1- cosθ)/sinθ now im stuck. help?

  • 2 years ago
  • 2 years ago

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  1. chaguanas Group Title
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    what did you do to the right? What is the assignment, are you proving one side equal to other, or solving equation?

    • 2 years ago
  2. rob1525 Group Title
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    what the original problem?

    • 2 years ago
  3. Algebraic! Group Title
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    yep, that works... keep going:)

    • 2 years ago
  4. zhenga2 Group Title
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    i am proving that both sides are equal and im only allowed to work on one side.

    • 2 years ago
  5. Algebraic! Group Title
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    \[\frac{ 1-\cos \theta }{\sin \theta } =\frac {\sin \theta } {1+\cos \theta }\]

    • 2 years ago
  6. Algebraic! Group Title
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    oh, only allowed to work on the LHS?

    • 2 years ago
  7. zhenga2 Group Title
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    that is what i have so far but im stuck and what does LHS mean?

    • 2 years ago
  8. Algebraic! Group Title
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    left hand side

    • 2 years ago
  9. zhenga2 Group Title
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    i can work on either side but once i start working on one side im limited to that side.

    • 2 years ago
  10. Algebraic! Group Title
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    ok

    • 2 years ago
  11. Algebraic! Group Title
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    use sin^2 x + cos ^2 x =1

    • 2 years ago
  12. Algebraic! Group Title
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    sin^2 x = 1- cos^2 x = (1+cosx)(1-cosx)

    • 2 years ago
  13. rob1525 Group Title
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    multiply the top and bottom by cos(theta) .right, then use identities to simplify.

    • 2 years ago
  14. Algebraic! Group Title
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    sin x / (1-cosx) = (1+cosx)/sin x

    • 2 years ago
  15. Algebraic! Group Title
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    or vice versa:)

    • 2 years ago
  16. chaguanas Group Title
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    It requires algebraic manipulation. Continuing from Algebraic work above|dw:1352775535108:dw|

    • 2 years ago
  17. chaguanas Group Title
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    Above is left hand side only, continuing from Algebraic start. Working on lefthand side only.

    • 2 years ago
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