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Math Help - Verify this identity

  1. #1
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    Verify this identity

    \frac{ \frac{1}{cos x}}{ \ \frac{1}{cos x} - \frac{sin x}{cos x}} = sec x(sec x+ tan x)

    I want to do this without turning the 1/cos x into sec x or sin x/cos x into tan x in the initial problem. Thanks!!!
    Last edited by Godzilla; November 19th 2009 at 11:48 AM.
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  2. #2
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    Hello, Godzilla!

    \frac{\dfrac{1}{\cos x}}{\dfrac{1}{\cos x} - \dfrac{\sin x}{\cos x}} \:=\: \sec x(\sec x+ \tan x)

    I want to do this without turning the 1/cos x into sec x or sin x/cos x into tan x in the initial problem.
    . . Why not? . . . It's the fastest way!
    We have: . \frac{\sec x}{\sec x - \tan x}


    \text{Multiply by }\frac{\sec x + \tan x}{\sec x + \tan x}\!:\quad<br />
\frac{\sec x}{\sec x - \tan x}\:\cdot\:\frac{\sec x + \tan x}{\sec x + \tan x}


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


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    Well, if you insist . . .

    Multiply by \frac{\cos x}{\cos x}\!:\quad \frac{\cos x\left(\dfrac{1}{\cos x}\right)} {\cos x\left(\dfrac{1}{\cos x} - \dfrac{\sin x}{\cos x}\right)} \;\;=\;\;\frac{1}{1-\sin x}


    Multiply by \frac{1+\sin x}{1+\sin x}\!:\quad \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}


    . . . . . =\;\;\frac{1}{\cos x}\left(\frac{1+\sin x}{\cos x}\right) \;\;=\;\;\frac{1}{\cos x}\left(\frac{1}{\cos x} + \frac{\sin x}{\cos x}\right) \;\;=\;\;\sec x(\sec x + \tan x)

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