Find the value of sec [Cot^-1 (-6)]?

Find the value of sec [Cot^-1 (-6)]?

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  • 1 decade ago
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    sec [arccot (-6)]

    the question is: what is the secant of that arc whose cot is (-6)?

    thus let [arccot (-6)] = y

    then cot y = - 6

    now you have to rewrite sec y in terms of cot y, that is:

    sec²y = (1/cos²y) = (cos²y + sin²y)/cos²y =

    1 + tan²y = 1 + 1/cot²y = (cot²y + 1)/cot²y

    and therefore, being cot y = - 6,

    sec y = ±√ [(cot²y + 1)/cot²y] = ± √ {(-6)² + 1]/(-6)²} =

    ±√ [(36 + 1)/(36)] = ± (√37)/6

    as for the sign, being arccot range (0,π), and being cot y negative,

    the respective arc belongs to the 2nd quadrant (π/2 < y < π)

    and therefore sec y is negative:

    in conlcusion,

    sec [arccot (-6)] = - (√37)/6

    I hope it has been helpful

    Bye!

  • 1 decade ago

    //////////

    draw triangle

    coty=-6

    adj=-6

    opp=1

    hyp=sqrt37

    secy=sqrt37/-6

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