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kaisan

simplify the expression: (sinx-1)(tanx+secx) help pretty please

  • one year ago
  • one year ago

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  1. kaisan
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    ?

    • one year ago
  2. kaisan
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    should i convert the tan and sec to sin and cos?

    • one year ago
  3. jaersyn
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    yes

    • one year ago
  4. Argonx16
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    Always start by breaking up Tangent/Cosecant/Secant/Cotangent into their constituents.

    • one year ago
  5. lgbasallote
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    ahh..an interesting question >:))

    • one year ago
  6. bmp
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    That simplifies to -cosx, right? If someone could check :-).

    • one year ago
  7. kaisan
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    i'm confused on what to do next >.<' i got this: |dw:1334814256375:dw|

    • one year ago
  8. Argonx16
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    You're pretty close. Remember that Secant is 1/Cosine, not 1/Sine.

    • one year ago
  9. jaersyn
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    that would make the problem infinitely easier haha

    • one year ago
  10. kaisan
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    |dw:1334814461856:dw| so then so i just put them together?

    • one year ago
  11. Argonx16
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    Given \[(sin(x-1))(tanx+secx)\] Simplify RHS \[(\sin(x-1)) * ((sinx/cosx)+(1/cosx))\] Which if you add the RHS together you should get \[(\sin(x-1))(1+sinx/cosx)\] And then you multiply.

    • one year ago
  12. kaisan
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    i multiply straight across?

    • one year ago
  13. lgbasallote
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    i think it was just \[\large sinx - 1\]right?

    • one year ago
  14. kaisan
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    the first part is

    • one year ago
  15. lgbasallote
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    so it'll be \[\large \mathbf{(sinx-1)*\frac{sinx+1}{cosx}}\] \[\large \mathbf{\frac{(sinx-1)(sinx + 1)}{cosx}}\]

    • one year ago
  16. lgbasallote
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    i'll let you guys do the rest :P Hint: difference of two squares

    • one year ago
  17. Argonx16
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    Oooh. I see what I did. That makes sense :D

    • one year ago
  18. kaisan
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    i got it ty ^^

    • one year ago
  19. kaisan
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    no wait its supposed to be negative but i got positive

    • one year ago
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