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kaisan

  • 2 years ago

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

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  1. kaisan
    • 2 years ago
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    ?

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

  3. jaersyn
    • 2 years ago
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    yes

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

  5. lgbasallote
    • 2 years ago
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    ahh..an interesting question >:))

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

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

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

  9. jaersyn
    • 2 years ago
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    that would make the problem infinitely easier haha

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

  11. Argonx16
    • 2 years ago
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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.

  12. kaisan
    • 2 years ago
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    i multiply straight across?

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

  14. kaisan
    • 2 years ago
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    the first part is

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

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

  17. Argonx16
    • 2 years ago
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    Oooh. I see what I did. That makes sense :D

  18. kaisan
    • 2 years ago
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    i got it ty ^^

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

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