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Math Help - Trigonomtertic Identity help

  1. #1
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    Trigonomtertic Identity help

    I need help proving the following trig identity: ( x stands for theta)
    csc x -1 / cot x = cot x / csc x + 1
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by mathmike View Post
    I need help solving the following trig identity: ( x stands for theta)
    csc x -1 / cot x = cot x / csc x + 1
    solving or proving? and do you mean \displaystyle \frac {\csc x - 1}{\cot x} = \frac {\cot x}{\csc x + 1} ?

    Hint: change everything to sines and cosines, and show you can simplify one side to get the other.
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  3. #3
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    Hello, mathmike!

    Are familiar with this identity and its variations?

    . \csc^2\!\theta \:=\:\cot^2\!\theta + 1 \quad\Rightarrow\quad \csc^2\!\theta -1\:=\:\cot^2\!\theta \quad\Rightarrow \quad \csc^2\!\theta - \cot^2\!\theta \:=\:1


    Prove: . \dfrac{\csc x -1}{\cot x} \:=\: \dfrac{\cot x}{\csc x + 1}

    The right side is: . \dfrac{\cot x}{\csc x + 1}

    Multiply by \frac{\csc x - 1}{\csc x - 1}\!:\;\;\dfrac{\cot x}{\csc x + 1}\cdot\dfrac{\csc x - 1}{\csc x - 1} \;=\; \dfrac{\cot x(\csc x - 1)}{\csc^2\!x - 1}

    . . . . . . . . . . . . =\;\dfrac{\cot x(\csc x-1)}{\cot^2\!x} \;=\;\dfrac{\csc x - 1}{\cot x}
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