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Math Help - trigo indentity

  1. #1
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    trigo indentity

    pls help me with this question, been working on it for quite some time but still can't figure it out:

    Given cosecA+cotA=3, evaluate cosecA-cotA and cosA.
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  2. #2
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    \frac{1}{\sin A}+\frac{\cos A}{\sin A} = 3
    Thus,
    1+\cos A = 3\sin A
    Square,
    1+2\cos A + \cos^2 A = 3\sin^2 A
    Thus,
    1+2\cos A+\cos^2 A = 3(1-\cos^2 A)
    Solve for,
    \cos A.
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  3. #3
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     <br />
\csc(A)+\cot(A) = \frac{1}{\sin(A)}+\frac{\cos(A)}{\sin(A)}<br />

     <br />
= \frac{1+\cos(A)}{\sin(A)}<br />

     <br />
= \frac{(1+\cos(A))(1-\cos(A))}{\sin(A)(1-\cos(A))}<br />

     <br />
= \frac{1-\cos^{2}(A)}{\sin(A)(1-\cos(A))}<br />

     <br />
= \frac{\sin^{2}(A)}{\sin(A)(1-\cos(A))}<br />

     <br />
= \frac{\sin(A)}{1-\cos(A)}<br />

     <br />
= \frac{1}{\frac{1-\cos(A)}{\sin(A)}}<br />

     <br />
= \frac{1}{\frac{1}{\sin(A)}-\frac{\cos(A)}{\sin(A)}}<br />

     <br />
= \frac{1}{\csc(A)-\cot(A)}<br />

    I suspect it's cleaner if you remember relationships between the cosecant and cotangent, but I never do. I always have to look them up.

    Okay, you do cos(A).
    Last edited by TKHunny; August 5th 2007 at 08:23 PM.
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  4. #4
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    Hello, wadevala!

    Given: . \csc A + \cot A \:=\:3, .evaluate: . \csc A - \cot A

    Note that: . {\color{blue}A \neq 0,\,\pi,\,\cdots}
    We have: . \csc A + \cot A \:=\:3


    Multiply by \frac{\csc A - \cot A}{\csc A - \cot A}\!:\;\;\frac{\csc A + \cot A}{1}\cdot\frac{\csc A - \cot A}{\csc A - \cot A} \;=\;3

    . . \frac{\csc^2\!A - \cot^2\!A}{\csc A - \cot A} \:=\:3\quad\Rightarrow\quad\frac{1}{\csc A - \cot A} \:=\:3

    Therefore: . \csc A - \cot A \:=\:\frac{1}{3}


    Evaluate: \cos A

    We have: . \frac{1}{\sin A} + \frac{\cos A}{\sin A} \:=\:3\quad\Rightarrow\quad1 + \cos A \:=\:3\sin A

    Square: . (1 + \cos A)^2 \:=\:9\sin^2\!A\quad\Rightarrow\quad1 + 2\cos A + \cos^2\!A \:=\:9(1-\cos^2\!A)

    . . which simplifies to: . 5\cos^2A + \cos A - 4 \:=\:0\quad\Rightarrow\quad(\cos + 1)(5\cos A - 4) \:=\:0


    and we have: . \begin{array}{cccccccccc}\cos A + 1 & = & 0 & \Rightarrow & \cos A & = & \text{-}1 & \Rightarrow & A \: = \:\pi & \text{extraneous} \\<br />
6\cos A - 4 & = & 0 & \Rightarrow & \cos A & = & \frac{4}{5} & \Rightarrow & A \: = \:\cos^{\text{-}1}\!\left(\frac{4}{5}\right) & <br />
\end{array}

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  5. #5
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    thx for the different ways of going about solving the qns, i understand it much better now
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