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Math Help - Formal Proof Assistance Please

  1. #1
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    Formal Proof Assistance Please

    I am having problems with a Formal Trigonometric Proof, and I hope to seek assistance here.

    (cotx)/(cscx +1)= (tan x)(csc x)-tanx

    I was able to work the left side to the following. (Work not included up to the point unless requested due to an error on my behalf.

    ( cos x ) (1 -sin x) / (1 - sinx^2)

    Now at this point, I am able to use the Pythagorean Identity to make the denominator cosx^2

    Once at this step I am unable to continue because I seem to not understand what step is next.


    Personally proofs have been difficult for me and any tips that you could leave me here would be much appreciated.
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  2. #2
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    Hello, jd3005!

    Prove that: . \frac{\cot x}{\csc x +1} \:=\: \tan x\csc x -\tan x

    Factor the right side: . \tan x\csc x - \tan x \;=\;\tan x(\csc x - 1)

    \text{Multiply by }\frac{\csc x + 1}{\csc x + 1}\!:\quad\frac{\tan x(\csc x - 1)}{1}\cdot\frac{\csc x + 1}{\csc x + 1} \;=\;\frac{\tan x\overbrace{(\csc^2\!x-1)}^{\text{This is }\cot^2\!x}}{\csc x + 1}

    \text{Then we have: }\;\frac{\tan x\cot^2\!x}{\csc x + 1} \;=\;\frac{\overbrace{(\tan x\cot x)}^{\text{This is 1}}\cot x}{\csc x + 1} \;=\;\frac{\cot x}{\csc x + 1}

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  3. #3
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    Quote Originally Posted by jd3005 View Post
    I am having problems with a Formal Trigonometric Proof, and I hope to seek assistance here.

    (cotx)/(cscx +1)= (tan x)(csc x)-tanx

    I was able to work the left side to the following. (Work not included up to the point unless requested due to an error on my behalf.

    ( cos x ) (1 -sin x) / (1 - sinx^2) Mr F says: Substitute 1 - sinx^2 = cosx^2 and cancel the common factor of cos x and you get (1 - sin x)/cos x.

    Now at this point, I am able to use the Pythagorean Identity to make the denominator cosx^2

    Once at this step I am unable to continue because I seem to not understand what step is next.


    Personally proofs have been difficult for me and any tips that you could leave me here would be much appreciated.
    My advice is now to simplify the right hand side (not quite according to Hoyle but nevertheless ....). It's not hard to get (1 - sin x)/cos x. After which the LHS is plainly equal to the RHS.
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  4. #4
    Super Member bigwave's Avatar
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    Prove that:
    \frac{\cot{x}}{\csc{x} +1}= \tan{x}\csc{x}-tan{x}

    from the LHS

    \frac{\cot{x}(\csc{x}-1)}{(\csc{x}+1)(\csc{x}-1)}<br />
\Rightarrow<br />
\frac{\cot{x}(\csc{x}-1)}{\cot^2{x}}<br />
\Rightarrow<br />
\frac{\csc{x}-1}{\cot{x}}<br />
\Rightarrow<br />
\frac{(\csc{x}-1)(\tan{x})}{\cot{x}\tan{x}}

    \Rightarrow<br />
\frac{\tan{x}\csc{x}-\tan{x}}{1}

    from RHS probably would be easier
    Last edited by bigwave; January 4th 2010 at 06:39 PM. Reason: cleanup
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  5. #5
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    Much thanks for all of you, however I am just curious as to whether the cosecant, secant etc are converted in sines and cosines.

    I know that in some cases it is like this, but is that only to reduce substantially the amount of work required for the problem?

    And in bigwave's last operation multiplying by tan x, was that only to rationalize the denominator?
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  6. #6
    Super Member bigwave's Avatar
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    disquard
    Last edited by bigwave; January 5th 2010 at 03:17 PM. Reason: not
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