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anonymous

  • one year ago

Trig/ Pre Cal/identies Am I on the right path? Please do not give answer but guide me. Thank you. The problem \[ \frac{1}{\sin x + 1} + \frac{1}{\csc x +1} \] \[ \frac{\csc x + 1 + \sin x + 1}{( \sin x + 1)( \csc x + 1)}\] Am I on the right path?

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  1. anonymous
    • one year ago
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    it should be \[ \frac{1}{\sin x + 1} + \frac{1}{\csc x +1} = 1 \] \[ \frac{\csc x + 1 + \sin x + 1}{( \sin x + 1)( \csc x + 1)} = 1\]

  2. Nnesha
    • one year ago
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    well first write csc in terms of sin :-) that will be easy :-)

  3. Nnesha
    • one year ago
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    \[ \frac{1}{\sin x + 1} + \frac{1}{\rm \color{reD}{\frac{1}{sinx} +1}} = 1 \] like this solve red part first

  4. anonymous
    • one year ago
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    So \[ \frac{\frac{1}{sin x} x + 1 + \sin x + 1}{( \sin x + 1)( \frac{1}{sin x} x + 1)}\]

  5. Nnesha
    • one year ago
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    better to gget rid of fraction first :-)

  6. anonymous
    • one year ago
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    so do I times the numerator and denominator by sin??

  7. Nnesha
    • one year ago
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    \[ \frac{1}{\sin x + 1} + \frac{1}{\rm \color{reD}{\frac{1}{sinx} +1}} = 1 \] \[\large\rm \color{reD}{\frac{ 1 }{ \sin x } +1}\] multiply by sinx/1 do you know how to find common denominator ? just for the red part

  8. anonymous
    • one year ago
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    For the red part the common denominator is sin x and 1. For the red we have \[ 1+ \sin x \] corect?

  9. Nnesha
    • one year ago
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    for the red part |dw:1434223526447:dw| common denominator is sin x

  10. anonymous
    • one year ago
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    yes correct. I forgot to include the denominator

  11. Nnesha
    • one year ago
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    in other words multiply by sinx/1 |dw:1434223578712:dw| let me know if you have any question :-)

  12. anonymous
    • one year ago
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    Got it

  13. Nnesha
    • one year ago
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    alright move on

  14. Nnesha
    • one year ago
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    \[ \huge\rm \frac{1}{\sin x + 1} + \frac{1}{\rm \color{reD}{\frac{1 + sinx }{sinx} }} = 1 \] now change divison to multiplication \[\huge\rm \frac{ 1 }{ \frac{ 1+sinx }{ sinx } }\] multiply top with the reciprocal of the bottom fraction

  15. Nnesha
    • one year ago
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    for example \[\large\rm \frac{ a }{ \frac{ b }{ c } } = a \times \frac{ c }{ b }\]

  16. anonymous
    • one year ago
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    so is it \( \frac{1}{1} \)

  17. anonymous
    • one year ago
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    so \( \frac{sin }{1+ sin x} \)

  18. Nnesha
    • one year ago
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    yes right :-)

  19. Nnesha
    • one year ago
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    \[ \huge\rm \frac{1}{\sin x + 1} + \frac{sin x}{\rm \color{reD}{ sinx +1 }} = 1 \] 1 + sinx is same as sinx +1

  20. anonymous
    • one year ago
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    so we have \[ \frac{sin x}{sinx + 1} = 1 \] but that does not equal 1]

  21. Nnesha
    • one year ago
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    check numerator again that's not right

  22. anonymous
    • one year ago
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    I see... Right in front of my face... lol man these things make me see crosseye

  23. anonymous
    • one year ago
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    so we have \[ \frac{1+sin x}{sinx + 1}\ = 1 but that does not equal 1]

  24. Nnesha
    • one year ago
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    why not ??

  25. anonymous
    • one year ago
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    so we have \[ \frac{1+sin x}{sinx + 1} = 1 \]

  26. Nnesha
    • one year ago
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    1+ sin x is same as sinx+1

  27. anonymous
    • one year ago
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    That post before my last post was from me copying and pasting and I did not take the last part out.

  28. anonymous
    • one year ago
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    Thank you. I was working it all types of ways but all I had to do was work the bottom denominator . Thank you!!

  29. Nnesha
    • one year ago
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    my pleasure :-) gO_od job!! :-)

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