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britbrat4290

  • one year ago

Rewrite the expression in terms of the given function or functions. (sec x + csc x) (sin x + cos x) - 2 - cot x; tan x

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  1. Coolsector
    • one year ago
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    tanx is the given function ?

  2. britbrat4290
    • one year ago
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    yea

  3. britbrat4290
    • one year ago
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    thats how its written

  4. Coolsector
    • one year ago
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    lets first look at : (secx + cscx)(sinx+cosx)

  5. Coolsector
    • one year ago
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    and you should know that secx = 1/cosx and cscx = 1/sinx

  6. Coolsector
    • one year ago
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    so can you simplify it ?

  7. britbrat4290
    • one year ago
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    what is cscx its not on my calc?

  8. Coolsector
    • one year ago
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    i just wrote it :) cscx = 1/sinx

  9. britbrat4290
    • one year ago
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    wow sry loll

  10. Coolsector
    • one year ago
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    so can you do it now ?

  11. britbrat4290
    • one year ago
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    1 sec

  12. Coolsector
    • one year ago
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    ok

  13. britbrat4290
    • one year ago
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    i get sec(x) for that prob up there

  14. britbrat4290
    • one year ago
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    and im wrong bc thats not a answer to choose from

  15. Coolsector
    • one year ago
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    lets do it together since i got something else. (secx + cscx)(sinx+cosx) = secxsinx + secxcosx + cscxsinx +cscxcosx = tanx + 1 + 1 + cotx

  16. Coolsector
    • one year ago
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    so (secx + cscx)(sinx+cosx) = 2 + cotx + tanx now lets add it to the rest : (secx + cscx)(sinx+cosx) - 2 - cotx = 2 + cotx + tanx -2 -2 - cotx = tanx

  17. Coolsector
    • one year ago
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    ill be back in some minutes i hope it is enough though

  18. britbrat4290
    • one year ago
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    i must not be puttin it i mcalc right

  19. britbrat4290
    • one year ago
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    2tan x??

  20. britbrat4290
    • one year ago
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    @Coolsector

  21. Coolsector
    • one year ago
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    @britbrat4290 hey im sorry im back now

  22. Coolsector
    • one year ago
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    as i wrote i got tanx

  23. britbrat4290
    • one year ago
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    shoot, ty

  24. Coolsector
    • one year ago
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    yw .. say hey to the dolphin

  25. britbrat4290
    • one year ago
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    :)

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