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help!!!!!

  • one year ago

Use basic identities to simplify the expression. one divided by cotangent of theta to the second power. + sec θ cos θ

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  1. help!!!!!
    • one year ago
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    please help me! I am taking precalc online and don't understand a single thing about basic identities

  2. hartnn
    • one year ago
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    so its \(\huge \dfrac{1}{\cot^2 \theta + \sec \theta \cos \theta} \)

  3. hartnn
    • one year ago
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    firstly, use the identity that sec theta = 1/cos theta so, sec theta cos theta = 1 and do you know what is \(\large 1+\cot^2 \theta = ... ?\)

  4. help!!!!!
    • one year ago
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    only the cot^2theta is the denominator

  5. help!!!!!
    • one year ago
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    no I don't know, I am COMPLETELY lost in this subject

  6. hartnn
    • one year ago
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    okk., so anyways you can use sec theta cos theta = 1 \(\large \dfrac{1}{\cot^2 \theta }+1 \)

  7. hartnn
    • one year ago
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    now 1/ cot theta = tan theta so, \(\large \tan^2 \theta+1 \)

  8. hartnn
    • one year ago
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    and ther's a standard pythagorean identity that \(\large 1+\tan^2 \theta = \sec^2 \theta \) thats it! :)

  9. help!!!!!
    • one year ago
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    okay thank you!

  10. help!!!!!
    • one year ago
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    can you walk me through another please?

  11. hartnn
    • one year ago
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    sure

  12. help!!!!!
    • one year ago
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    Simplify the expression. quantity cosecant of x to the power of two times secant of x to the power of two divided by quantity secant of x to the power of two plus cosecant of x to the power of two.

  13. help!!!!!
    • one year ago
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    |dw:1389469428352:dw|

  14. help!!!!!
    • one year ago
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    yes

  15. hartnn
    • one year ago
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    \(\large \dfrac{\csc^2 x \sec^2 x }{\sec^2 x + \csc^2 x }\)

  16. hartnn
    • one year ago
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    ok, we use the identity that csc x = 1/ sin x sec x = 1/cos x

  17. help!!!!!
    • one year ago
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    okay

  18. hartnn
    • one year ago
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    \(\large \dfrac{\csc^2 x \sec^2 x }{\sec^2 x + \csc^2 x } = \large \dfrac{\sin^2 x \cos^2 x }{\sin^2 x + \cos^2 x }\)

  19. hartnn
    • one year ago
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    now we can use the famous pythagorean identity \(\Large \color{red}{\sin^2x + \cos^2 x =1 }\)

  20. hartnn
    • one year ago
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    what remains is your answer! :)

  21. help!!!!!
    • one year ago
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    so the answer is just 1? or am I wrong?

  22. hartnn
    • one year ago
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    denominator = 1 yes what remains is numerator sin^2 x cos^2 x

  23. hartnn
    • one year ago
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    you have choices for this too ?

  24. help!!!!!
    • one year ago
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    yes I do the choices are: sin^2x cos^2x -1 1

  25. help!!!!!
    • one year ago
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    im sorry im so stupid. im usually really good at math but precalc just makes no sense to me

  26. hartnn
    • one year ago
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    oops, i did a simplification error \(\large \dfrac{\csc^2 x \sec^2 x }{\sec^2 x + \csc^2 x } = \large \dfrac{1 }{\sin^2 x + \cos^2 x } = 1\)

  27. help!!!!!
    • one year ago
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    okay and i have a question. next year I planned on taking ap stat. is that anything like precalc because i hope not. is it easier or harder?

  28. hartnn
    • one year ago
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    if you practice less, math is hard. if you practice more, math is easy. irrespective of any topic/subject

  29. help!!!!!
    • one year ago
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    well are the concepts of ap stat very similar to or very different to those of pre-calculus?

  30. hartnn
    • one year ago
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    statistics is a different branch of math altogether, so very different.

  31. help!!!!!
    • one year ago
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    i just think I'm not good at online class taking because I learn better from an actual person rather than a computer screen. I need the physical classroom environment and a teacher to give me instruction.

  32. help!!!!!
    • one year ago
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    what kind of problems does ap stat entail? word problems? graphs?

  33. hartnn
    • one year ago
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    both.

  34. help!!!!!
    • one year ago
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    oh okay thank you soooooo much! is there anything else that i should know?

  35. hartnn
    • one year ago
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    try to take up easy questions before going to harder ones. if you get stuck anytime, OpenStudy is always Open for you :)

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