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Math Help - trig proof help

  1. #1
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    trig proof help

    Show that;

     \frac{cotx}{cosecx-1} -\frac{cosx}{1+sinx} = 2tanx
     \frac {\frac{cosx}{sinx}}{ \frac{1}{sinx}-1} - \frac{cosx}{1+sinx}

     \frac{ \frac{cosx}{sinx}}{ \frac{1-sinx}{sinx} }- \frac{cosx}{1+sinx}

     \frac{cosx}{sinx} \times \frac{sinx}{1-sinx} = \frac{cosx sinx}{sinx-sin^2x}

    Than can I cancel out the sinx's? so I get;

     \frac{cosx}{-sinx} - \frac{cosx}{1+sinx} = 2tanx

    Than I am not sure where to go from here, any suggestions appreciated.

    Thank you!
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  2. #2
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    Quote Originally Posted by Tweety View Post
     \frac {\frac{cosx}{sinx}}{ \frac{1}{sinx}-1} - \frac{cosx}{1+sinx}

     \frac{ \frac{cosx}{sinx}}{ \frac{1-sinx}{sinx} }- \frac{cosx}{1+sinx} ***

     \frac{cosx}{sinx} \times \frac{sinx}{1-sinx} = \frac{cosx sinx}{sinx-sin^2x}

    Than can I cancel out the sinx's? so I get;

     \frac{cosx}{-sinx} - \frac{cosx}{1+sinx} = 2tanx

    Than I am not sure where to go from here, any suggestions appreciated.

    Thank you!
    HI

    *** I will continue from here .

    \frac{\cos x}{1-\sin x}-\frac{\cos x}{1+\sin x}

    =\frac{\cos x(1+\sin x-\cos x(1-\sin x))}{1-\sin^2 x}

    =\frac{\cos x=\cos x\sin x-\cos x+\cos x\sin x}{\cos^2 x}

    =\frac{2\cos x\sin x}{\cos^2 x}

    =2\frac{\sin x}{\cos x}

     <br />
=2\tan x<br />
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  3. #3
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    Quote Originally Posted by mathaddict View Post
    HI

    *** I will continue from here .

    \frac{\cos x}{1-\sin x}-\frac{\cos x}{1+\sin x}

    =\frac{\cos x(1+\sin x-\cos x(1-\sin x))}{1-\sin^2 x}

    =\frac{\cos x=\cos x\sin x-\cos x+\cos x\sin x}{\cos^2 x}

    =\frac{2\cos x\sin x}{\cos^2 x}

    =2\frac{\sin x}{\cos x}

     <br />
=2\tan x<br />
    hey thanks for that but I am not sure how you got 1-sinx as the denomninator?


    Could you explain ,

    thanks
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  4. #4
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    \frac{ \frac{cosx}{sinx}}{ \frac{1-sinx}{sinx} }- \frac{cosx}{1+sinx}<br />

    When at this point, the sinx as denominators (in the first fraction) cancel each other out leaving the 1-sin(x) as the overall denominator
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  5. #5
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    Quote Originally Posted by Beard View Post
    \frac{ \frac{cosx}{sinx}}{ \frac{1-sinx}{sinx} }- \frac{cosx}{1+sinx}<br />

    When at this point, the sinx as denominators (in the first fraction) cancel each other out leaving the 1-sin(x) as the overall denominator
    Thank you, Also is that a general rule that when you have one fraction on top of another fraction and thier denominators are the same you can cancel them?
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  6. #6
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    Hello Tweety
    Quote Originally Posted by Tweety View Post
    Thank you, Also is that a general rule that when you have one fraction on top of another fraction and thier denominators are the same you can cancel them?
    Yes. When you have something like
    \frac{\left(\dfrac{A}{B}\right)}{\left(\dfrac{C}{B  }\right)}
    you can multiply top-and-bottom of the main fraction by B:
    \frac{\left(\dfrac{A}{B}\right)\times B}{\left(\dfrac{C}{B}\right)\times B}
    which then cancels each of the 'B' denominators to leave you with
    \frac{A}{C}
    Grandad
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  7. #7
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    Quote Originally Posted by Beard View Post
    \frac{ \frac{cosx}{sinx}}{ \frac{1-sinx}{sinx} }- \frac{cosx}{1+sinx}<br />

    When at this point, the sinx as denominators (in the first fraction) cancel each other out leaving the 1-sin(x) as the overall denominator
    Quote Originally Posted by Grandad View Post
    Hello TweetyYes. When you have something like
    \frac{\left(\dfrac{A}{B}\right)}{\left(\dfrac{C}{B  }\right)}
    you can multiply top-and-bottom of the main fraction by B:
    \frac{\left(\dfrac{A}{B}\right)\times B}{\left(\dfrac{C}{B}\right)\times B}
    which then cancels each of the 'B' denominators to leave you with
    \frac{A}{C}
    Grandad
    Thank you.
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