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Math Help - Another trig identity

  1. #1
    Newbie
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    Feb 2011
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    Another trig identity

    Very Helpful thank you very much.

    There is one more I am struggling with however.

    sec^2x cscx = cscx +secx tanx

    Any ideas?
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  2. #2
    Super Member

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    Lexington, MA (USA)
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    Hello, Jkeeb!

    We can prove it in both directions.


    \sec^2\!x\csc x \:=\: \csc x + \sec x\tan x

    The left side is: . \sec^2\!x\csc x

    . . . . . . . . . . =\;(1 + \tan^2\!x)\csc x

    . . . . . . . . . . =\;\csc x + \tan^2\!x\csc x

    . . . . . . . . . . \displaystyle =\;\csc x + \frac{\sin^2\!x}{\cos^2\!x}\!\cdot\!\frac{1}{\sin x}

    . . . . . . . . . . \displaystyle =\;\csc x + \frac{1}{\cos x}\!\cdot\!\frac{\sin x}{\cos x}

    . . . . . . . . . . =\;\csc x + \sec x\tan x



    The right side is: . \csc x + \sec x\tan x

    . . . . . . . . . . . \displaystyle =\;\frac{1}{\sin x} + \frac{1}{\cos x}\!\cdot\!\frac{\sin x}{\cos x}

    . . . . . . . . . . . \displaystyle =\;\frac{\overbrace{\cos^2\!x + \sin^2\!x}^{\text{This is 1}}}{\cos^2\!x\sin x}

    . . . . . . . . . . . \displaystyle =\;\frac{1}{\cos^2\!x\sin x}

    . . . . . . . . . . . \displaystyle =\;\frac{1}{\cos^2\!x}\!\cdot\!\frac{1}{\sin x}

    . . . . . . . . . . . =\; \sec^2\!x\csc x

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