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Math Help - need help very urgent

  1. #1
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    need help very urgent

    can show mi the step...thanx

    if sin x=-0.6 and tan x = -0.75,then sec x is

    1)-1.25
    2)1.25
    3)0.80
    4)-1.35
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  2. #2
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    Quote Originally Posted by cheesepie
    can show mi the step...thanx

    if sin x=-0.6 and tan x = -0.75,then sec x is

    1)-1.25
    2)1.25
    3)0.80
    4)-1.35
    Hello,

    1. You know that \tan(x)=\frac{\sin(x)}{\cos(x)} and

    \sec(x)=\frac{1}{\cos(x)} so

    \sec(x)=\frac{\tan(x)}{\sin(x)}

    Therefore: \frac{-\frac{3}{4}}{-\frac{3}{5}}=\frac{5}{4}

    Thus the answer is 2)

    Greetings

    EB
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  3. #3
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    Quote Originally Posted by cheesepie
    can show mi the step...thanx

    if sin x=-0.6 and tan x = -0.75,then sec x is

    1)-1.25
    2)1.25
    3)0.80
    4)-1.35
    \tan(x)=\sin(x)/\cos(x)=\sin(x)\sec(x),

    so:

    <br />
\sec(x)=\tan(x)/\sin(x)=(-0.75)/(-0.6)=1.25<br />

    RonL
    Last edited by CaptainBlack; May 11th 2006 at 08:07 AM.
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  4. #4
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    Quote Originally Posted by CaptainBlack
    \tan(x)=\sin(x)/\cos(x)=\sin(x)\sec(x),

    so:

    <br />
\sec(x)=\tan(x)/\sin(x)=-0.75/0.6=-1.25<br />

    RonL
    Hello,

    I'm a little bit confused, because I read a sin x = -0.6 in the problem of cheesepie.

    Greetings

    EB
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  5. #5
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    can u show mi the step of this question ? thanx

    if cos x=0.6 and tan x is negative ,then csc x is


    1)1.25
    2)-1.25
    3)0.80
    4)0.75
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  6. #6
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    Quote Originally Posted by earboth
    Hello,

    I'm a little bit confused, because I read a sin x = -0.6 in the problem of cheesepie.

    Greetings

    EB
    Sorry I misread the value

    RonL
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  7. #7
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    Quote Originally Posted by cheesepie
    can u show mi the step of this question ? thanx
    if cos x=0.6 and tan x is negative ,then csc x is
    1)1.25
    2)-1.25
    3)0.80
    4)0.75
    Hello,

    This problem is a tricky one, because you didn't give a value for tan(x). To be able to calculate a value I take the value from the previous problem: tan(x)=-.75

    1. You know that \cot(x)={\cos(x) \over \sin(x)} and

    \csc(x)={1 \over \sin(x)} and \tan(x)={\sin(x) \over \cos(x)} So you get:

    \csc(x)={{\cos(x) \over \sin(x)} \over \cos(x)}} = {{1 \over \tan(x)} \over \cos(x)}} = {{1 \over -.75} \over .6}} = -{20 \over 9} \approx -2.22

    This answer doesn't exist. So I presume, that my asumption about the value of tan(x) was incorrect. But maybe you can use my explanations when you plug in the right value for tan(x).

    Greetings

    EB
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  8. #8
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by cheesepie
    can u show mi the step of this question ? thanx

    if cos x=0.6 and tan x is negative ,then csc x is


    1)1.25
    2)-1.25
    3)0.80
    4)0.75
    We know that cos x = 0.6
    Thus: sin \,x = \pm \sqrt{1-0.6^2}=\pm 0.8

    Since tan x is negative, sin x = -0.8

    Thus csc x = 1/sin x = 1/-0.8 = -1.25

    -Dan
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