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Math Help - Trig identity

  1. #1
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    Trig identity

    hi i really need help with this homework trig identity it's really angering me. please tell me how you would start it off to get me started thanks.

    (secx + tanx) / (secx - tanx) = (1 + 2sinx + sin(x)^2) / cos(x)^2

    thanks
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  2. #2
    A Plied Mathematician
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    I would definitely start off by writing everything in terms of sines and cosines. Where does that take you?
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  3. #3
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    Hello, TacticalPro!

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

    Start with the left side; multiply by \dfrac{\sec x + \tan x}{\sec x + \tan x}

    . . \displaystyle \frac{\sec x + \tan x}{\sec x - \tan x}\cdot\frac{\sec x + \tan x}{\sec x + \tan x}


    . . \displaystyle =\;\frac{(\sec x + \tan x)^2}{\underbrace{\sec^2\!x - \tan^2\!x}_{\text{This is 1}}}


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


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


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

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