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GOODMAN

  • 2 years ago

Verify that this is an identity. sec(beta)+csc(beta) All over: 1+tan(beta) is equal to=csc(beta)

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  1. GOODMAN
    • 2 years ago
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    |dw:1337050670476:dw|

  2. lgbasallote
    • 2 years ago
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    hmm maybe rewrite tan B as \(\frac{\sin B}{\cos B}\)?

  3. Callisto
    • 2 years ago
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    \[\frac{sec\beta + csc\beta}{1+tan\beta} = \frac{\frac{sin\beta+cos\beta}{sin\beta cos\beta}}{\frac{sin\beta + cos\beta}{cos\beta}} = \frac{1}{sin\beta}=csc\beta\]

  4. GOODMAN
    • 2 years ago
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    lol..thats soo small.

  5. Callisto
    • 2 years ago
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    How can I make it bigger?

  6. lgbasallote
    • 2 years ago
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    \[\frac{\frac{1}{\cos B} + \frac{1}{\sin B}}{\frac{\cos B + \sin B}{\cos B}}\] \[\frac{\frac{\sin B + \cos B}{\sin B \cos B}}{\frac{\cos B + \sin B}{\cos B}}\] \[\frac{\frac{1}{\sin B \cos B}}{\frac{1}{\cos B}}\]

  7. lgbasallote
    • 2 years ago
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    ugh @Callisto said it :P

  8. GOODMAN
    • 2 years ago
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    No..wait..im still a little confused, how did you guys split it?

  9. lgbasallote
    • 2 years ago
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    \[\large \frac{1}{\sin B} = \csc B\]

  10. lgbasallote
    • 2 years ago
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    split what?

  11. GOODMAN
    • 2 years ago
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    Yes..i know that, i know the basic identities.

  12. GOODMAN
    • 2 years ago
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    I mean, like the second and third step you did.

  13. lgbasallote
    • 2 years ago
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    \[\large \sec B + \csc B = \frac{1}{\cos B} + \frac{1}{\sin B}\] LCD \[\frac{\sin B + \cos B}{\sin B \cos B}\]

  14. Callisto
    • 2 years ago
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    \[\frac{1}{cos \beta} + \frac{1}{sin\beta} = \frac{1(sin\beta)}{cos \beta sin\beta} + \frac{1(cos \beta)}{sin\beta cos \beta} =\frac{sin\beta +cos\beta}{sin\beta cos\beta} \]

  15. lgbasallote
    • 2 years ago
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    \[1 + \tan B = \frac{\cos B}{\cos B} + \frac{\sin B}{\cos B}\]

  16. lgbasallote
    • 2 years ago
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    \[\frac{\sin B + \cos B}{\cos B}\]

  17. Callisto
    • 2 years ago
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    \[1+tan\beta = 1+ \frac{sin\beta}{cos\beta} = \frac{cos\beta}{cos\beta} +\frac{sin\beta}{cos\beta} = \frac{sin\beta+cos \beta}{cos\beta} \]

  18. lgbasallote
    • 2 years ago
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    the \(\sin B + \cos B\) cancels

  19. GOODMAN
    • 2 years ago
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    Oh..okayy...basically i had to lay out the fractions? Arghh this stuff is weird.

  20. lgbasallote
    • 2 years ago
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    \[\frac{\frac{1}{\sin B \cos B}}{\frac{1}{\cos B}}\]\[\frac{\cos B}{\sin B \cos B}\]

  21. Callisto
    • 2 years ago
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    When you make it like that, it's easier to simplify~ Maths is weird in nature... I think :S

  22. GOODMAN
    • 2 years ago
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    Okay, I think i got it from here, thanks :D

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