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pvs285

  • 2 years ago

(csc x + 1)(csc x - 1)

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  1. NoelGreco
    • 2 years ago
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    FOIL it and apply trig identity

  2. itsmylife
    • 2 years ago
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    so its \[(\csc x)^2-1=cot^22x\]

  3. NoelGreco
    • 2 years ago
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    yep

  4. pvs285
    • 2 years ago
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    @itsmylife how did you get that

  5. itsmylife
    • 2 years ago
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    trig identity

  6. pvs285
    • 2 years ago
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    how

  7. pvs285
    • 2 years ago
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    can you show how you got that plz

  8. itsmylife
    • 2 years ago
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    pythagorean identity

  9. NoelGreco
    • 2 years ago
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    There are three Pythagorean trig identities that you simply must memorize.

  10. itsmylife
    • 2 years ago
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    go here @pvs285 http://www.sosmath.com/trig/Trig5/trig5/trig5.html

  11. pvs285
    • 2 years ago
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    can you just show the work to answer this question plz?

  12. surjithayer
    • 2 years ago
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    \[\left( \csc x+1 \right)\left( \csc x-1 \right)=\csc ^{2}x-1=\cot ^{2}x\]

  13. itsmylife
    • 2 years ago
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    \[\sin^2x+\cos^2x=1,\frac{ \sin^2 x}{ \sin^2x }+\frac{ \cos^2x }{ \sin^2x }=\frac{ 1 }{ \sin^2x }, 1+\cot^2x=cosec^2x\]

  14. NoelGreco
    • 2 years ago
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    It is against the policies of this site to simply provide answers.

  15. pvs285
    • 2 years ago
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    huh u said it was cot^2

  16. pvs285
    • 2 years ago
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    ik i whant you to explain it to me so i uderstand

  17. itsmylife
    • 2 years ago
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    it might be a mistake ;)

  18. pvs285
    • 2 years ago
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    huh

  19. pvs285
    • 2 years ago
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    pls help

  20. pvs285
    • 2 years ago
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    plz

  21. surjithayer
    • 2 years ago
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    \[\csc ^{2}x-\cot ^{2}=1,or \csc ^{2}x-1=\cot ^{2}x\] It is an identity.

  22. pvs285
    • 2 years ago
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    so there is no work involved?

  23. NoelGreco
    • 2 years ago
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    pvs: it appears you are in over your head. If you have a question such as this, and can't pick up on the idea of a trig identity you might want to slow down and back up.

  24. ybarrap
    • 2 years ago
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    $$ \tt \bf \Large(\csc x + 1)(\csc x -1)\\ \Large=\csc^2x-\csc x +\csc x -1\\ \Large=\csc^2x-1\\\\ \Large=\left(\dfrac{1}{\sin x}\right)^2-1\\ \Large=\dfrac{1}{\sin^2x}-1\\ \Large=\dfrac{1}{\sin^2x}-\dfrac{\sin^2x}{\sin^2x}\\\\ \Large=\dfrac{1-\sin^2x}{\cos^2x}\\\\ \Large=\dfrac{\cos^2x}{\sin^2x}\\\\ \Large=\cot^2x $$ Please let me kow if you have any questions. Good luck.

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