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Math Help - FInd exact value of the expression Sec(arctan -5/12)

  1. #1
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    FInd exact value of the expression Sec(arctan -5/12)

    Hey guys i want to know if i am correct here

    let y=arctan(-5/12)

    sec y= 13/12 or 65 degrees

    is this correct
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  2. #2
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    Yep, except you might not want the 65 degrees as your answer as that is obviously not an exact answer and is not even equivalent to your expression.

    \sec\left[ \arctan \left( -\frac{5}{12}\right) \right] should be unitless.
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  3. #3
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    Quote Originally Posted by NoAsherelol View Post
    Hey guys i want to know if i am correct here

    let y=arctan(-5/12)

    sec y= 13/12 or 65 degrees

    is this correct
    The correct answer is NOT an angle .....

    \sec\left[ \arctan \left( -\frac{5}{12}\right) \right] = \frac{13}{12} OR {\color{red}-} \frac{13}{12}.
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    Quote Originally Posted by NoAsherelol View Post
    Hey guys i want to know if i am correct here

    let y=arctan(-5/12)

    sec y= 13/12 or 65 degrees

    is this correct
    \sec\left(\arctan(x)\right)=\sqrt{x^2+1}
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  5. #5
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    Quote Originally Posted by NoAsherelol View Post
    Hey guys i want to know if i am correct here

    let y=arctan(-5/12)

    sec y= 13/12 or 65 degrees

    is this correct
    If y = arctan(-5 /12),
    then, tan(y) = -5/ 12
    So, y is an angle whose tan value is -5 /12.

    tan = (opposite side)/(adjacent side) = opp/adj

    So, for angle y above,
    opp = -5
    adj = 12
    The hypotenuse---hyp---then is sqrt[(-5)^2 +(12)^2] = 13

    sec = hyp/adj
    So, sec(y) = 13/12 -----you got that right.

    -----------------------

    If y = arctan(5/(-12)),
    Then, opp = 5, adj = -12, hyp = 13

    sec(y) = 13/(-12) or -(13/12)
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