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Math Help - Solution required urgently

  1. #1
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    Solution required urgently

    Hi,




    • Find x from each of the following equations.
    1.1)x*tan^2(45)*cos(60) = sin(60)*cot(30)

    1.2)(1+x)*tan(45)*cosec^2(45) = sin(30)*sec^2(45)

    1.3)X*cos(60)–sin(30) = x*tan(30)*cot(60)


    1.sec(120)



    2.sec(350)



    3.tan(40)



    Request to respond as soon as possible.


    Best Regards,
    Lalit Chugh
    Last edited by MathGuru; June 19th 2006 at 11:56 AM. Reason: should not be an attachemnt, too simple
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  2. #2
    Super Member malaygoel's Avatar
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    Quote Originally Posted by lalitchugh
    Hi,




    • Find x from each of the following equations.
    1.1)x*tan^2(45)*cos(60) = sin(60)*cot(30)

    1.2)(1+x)*tan(45)*cosec^2(45) = sin(30)*sec^2(45)

    1.3)X*cos(60)–sin(30) = x*tan(30)*cot(60)


    1.sec(120)



    2.sec(350)



    3.tan(40)



    Request to respond as soon as possible.
    Best Regards,
    Lalit Chugh
    (1.1) xtan^{2}45cos60 = sin60cot30
    tan45=1
    cos60=\frac{1}{2}
    sin60=\frac{\sqrt{3}}{2}
    cot30=\sqrt{3}
    Hence
    x=\frac{\frac{\sqrt{3}}{2}*\sqrt{3}}{1^2*\frac{1}{  2}}
    x=3
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  3. #3
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    Hello, lalitchugh!

    You're expected to know the trig values for 30°, 60°, and 45°.
    Once you plug in the values, the rest is Arithmetic and Algebra I . . .

    1.2)\;\;(1+x)\cdot\tan(45^o)\cdot\csc^2(45^o) \;= \;\sin(30^o)\cdot\sec^2(45^o)
    We have: . (1 + x)(1)(\sqrt{2})^2\;=\;\left(\frac{1}{2}\right)( \sqrt{2})^2

    Then: . (1 + x)\cdot 2\;=\;1\quad\Rightarrow\quad 1 + x \:=\:\frac{1}{2}\quad\Rightarrow\quad x\,=\,-\frac{1}{2}


    1.3)\;\;x\cdot\cos(60^o) - \sin(30^o) \;= \;x\cdot\tan(30^o)\cdot\cot(60^o)
    We have: . x\left(\frac{1}{2}\right) - \frac{1}{2} \;= \;x\left(\frac{1}{\sqrt{3}}\right)\left(\frac{1}{ \sqrt{3}}\right)

    Then: . \frac{1}{2}x - \frac{1}{2}\;=\;\frac{1}{3}x\quad\Rightarrow\quad \frac{1}{6}x\,=\, \frac{1}{2}\quad\Rightarrow\quad x = 3
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  4. #4
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    Solution required urgently

    Hi,

    Thanks for the reply.

    Can you pls advice how we can evaluate

    1. tan 40

    2. sec 350

    I tried a lot but could not solve these.


    Best Regards,
    Lalit Chugh
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  5. #5
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    Quote Originally Posted by lalitchugh
    Hi,

    Thanks for the reply.

    Can you pls advice how we can evaluate

    1. tan 40

    2. sec 350

    I tried a lot but could not solve these.


    Best Regards,
    Lalit Chugh
    I do not see a simple way to do it. It leads to a cubic equation which is probably not what you are looking for.
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  6. #6
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    Solution required urgently

    Hi,

    Many thanks for the update.


    Cheers
    Lalit Chugh
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  7. #7
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    Post hwy buddy

    Quote Originally Posted by lalitchugh
    Hi,

    Thanks for the reply.

    Can you pls advice how we can evaluate

    1. tan 40

    2. sec 350

    I tried a lot but could not solve these.


    Best Regards,
    Lalit Chugh


    HEY BUDDY THESE VALUES CAN BE LOOKED UPON FROM TRIGNOMETRIC TABLES ONLY THERE IS NO SOLUTIONF FOR THEM
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  8. #8
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    Quote Originally Posted by killer baby
    HEY BUDDY THESE VALUES CAN BE LOOKED UPON FROM TRIGNOMETRIC TABLES ONLY THERE IS NO SOLUTIONF FOR THEM
    There is a solution for them, it is only too complicated.

    Do not talk with caps on!
    -=USER WARNED=-
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