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Math Help - Proving Trig Identities?

  1. #1
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    Proving Trig Identities?

    I cannot understand this at all. how do i prove:

    1-sinx/1+sinx = (tanx-secx)^2
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  2. #2
    MHF Contributor red_dog's Avatar
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    (\tan x-\sec x)^2=\left(\frac{\sin x}{\cos x}-\frac{1}{\cos x}\right)^2=\left(\frac{\sin x-1}{\cos x}\right)=

    =\frac{(1-\sin x)^2}{\cos ^2x}=\frac{(1-\sin x)^2}{1-\sin^2x}=\frac{(1-\sin x)^2}{(1-\sin x)(1+\sin x)}=\frac{1-\sin x}{1+\sin x}
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  3. #3
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    thank you so much! this proving trig identities stuff is killing me. i cannot understand it.
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  4. #4
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    Hello, ccarnessali!

    We can go the other way, too . . .


    \frac{1-\sin x}{1+\sin x} \:=\: (\tan x-\sec x)^2
    Multiply the left side by: \frac{1-\sin x}{1-\sin x}

    . . \frac{1-\sin x}{1+\sin x}\cdot {\color{blue}\frac{1-\sin x}{1-\sin x}}


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


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


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


    . . = \;\left(\frac{1}{\cos x}\right)^2 - 2\left(\frac{1}{\cos x}\right)\left(\frac{\sin x}{\cos x}\right) + \left(\frac{\sin x}{\cos x}\right)^2


    . . = \;\sec^2\!x - 2\sec x\tan x + \tan^2\!x


    . . = \;(\sec x - \tan x)^2

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