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anonymous

  • one year ago

VERIFY THE IDENTITY: 1/csc(x-1) + 1/csc(x+1)=2/csc(x)-sin(x)

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  1. freckles
    • one year ago
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    question is it really: \[\frac{1}{\csc(x-1)}+\frac{1}{\csc(x+1)}=\frac{2}{\csc(x)}-\sin(x)\]

  2. anonymous
    • one year ago
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    close, only it says 2/csc(x)-sin(x) not sure if that is the same thing or not?

  3. freckles
    • one year ago
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    isn't that what I typed for the right hand side?

  4. freckles
    • one year ago
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    did you mean to write 2/(csc(x)-sin(x)) instead?

  5. anonymous
    • one year ago
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    yes

  6. freckles
    • one year ago
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    \[\frac{1}{\csc(x-1)}+\frac{1}{\csc(x+1)}=\frac{2}{\csc(x)-\sin(x)}\]

  7. anonymous
    • one year ago
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    yes that is correct

  8. freckles
    • one year ago
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    that doesn't seem like an identity

  9. freckles
    • one year ago
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    try inputting pi/6

  10. freckles
    • one year ago
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    both sides aren't the same

  11. anonymous
    • one year ago
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    Ok that is what I kept coming up with, I just wanted to verify that it was not an identity, thank you!

  12. freckles
    • one year ago
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    did you mean to write \[\frac{1}{\csc(x)-1}+\frac{1}{\csc(x)+1}=\frac{2}{\csc(x)-\sin(x)}\] this is an identity

  13. anonymous
    • one year ago
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    I cannot figure out how it is an identity, however....

  14. freckles
    • one year ago
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    so is that not what you are trying to prove is an identity ? it really is the first equation you wrote?

  15. freckles
    • one year ago
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    If you really meant to write that last thing I wrote, try combining the fractions on the left hand side first

  16. anonymous
    • one year ago
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    I think I am typing it wrong, this is exactly how it is written on the worksheet: \[\frac{ 1 }{ \csc x-1 }+\frac{ 1 }{ \csc x+1 }=\frac{ 2 }{ \csc x-\sin x }\]

  17. freckles
    • one year ago
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    then all you have to do is divide both top and bottom by csc(x)

  18. freckles
    • one year ago
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    2 steps really: combine fractions divide top and bot by csc(x)

  19. anonymous
    • one year ago
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    How can you combine the fractions when the denominator are not the same? I am not sure how to make them the same

  20. UnkleRhaukus
    • one year ago
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    \[LHS=\frac{1}{\csc(x)-1}+\frac{1}{\csc(x)+1}\\ =\frac{1}{\csc(x)-1}\times\frac{\csc(x)+1}{\csc(x)+1}+\frac{1}{\csc(x)+1}\times\frac{\csc(x)-1}{\csc(x)-1}\\ =\frac{2\csc(x)}{\csc^2(x)-1}\\ =\] now divide both numerator and denominator by csc

  21. anonymous
    • one year ago
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    Like this? \[\frac{ 2 }{ \csc x-1}\]

  22. UnkleRhaukus
    • one year ago
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    \[\frac{ 2 }{ \csc x-\tfrac1{\csc(x)}}\]

  23. anonymous
    • one year ago
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    ohh of course. I understand now. Thank you so much!

  24. UnkleRhaukus
    • one year ago
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    all clear? do you understand how we combined those fractions at the start?

  25. anonymous
    • one year ago
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    yes, I kept trying to do it that way but for some reason I was doing the algebra wrong....I need to practice more

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