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Math Help - Identities Crisis

  1. #1
    Junior Member Freaky-Person's Avatar
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    Identities Crisis

    1/1+Sin = Sec^2 - Tan/Cos
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  2. #2
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    Quote Originally Posted by Freaky-Person View Post
    1/1+Sin = Sec^2 - Tan/Cos
    I presume,
    \frac{1}{1+\sin x}=\sec^2x - \frac{\tan x}{\cos x}
    Work one left hand side,
    \frac{1}{1+\sin x}\cdot \frac{1-\sin x}{1-\sin x}=\frac{1-\sin x}{1-\sin^2 x}=\frac{1-\sin x}{\cos^2 x}
    Split fraction,
    \frac{1}{\cos^2 x}-\frac{\sin x}{\cos ^2 x}=\sec^2 x-\frac{\tan x}{\cos x}
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  3. #3
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    Hello, Freaky-Person!

    Another approach . . .


    \frac{1}{1 + \sin x} \:= \:\sec^2\!x - \frac{\tan x}{\cos x}

    Start with the right side . . .

    \sec^2\!x - \frac{\tan x}{\cos x} \;=\;\frac{1}{\cos^2\!x} - \frac{\frac{\sin x}{\cos x}}{\cos x}\;=\;\frac{1}{\cos^2x} - \frac{\sin x}{\cos^2\!x} \;=\;\frac{1 - \sin x}{\cos^2\!x}


    Multiply top and bottom by (1 + \sin x)\!:

    . . \frac{1 + \sin x}{1 + \sin x}\cdot\frac{1 - \sin x}{\cos^2\!x} \;=\; \frac{1 - \sin^2\!x}{(1 + \sin x)\cos^2\!x} \:=\:\frac{\cos^2\!x}{(1 + \sin x)\cos^2\!x} \:=\:\frac{1}{1 + \sin x}

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