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Math Help - identities

  1. #1
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    identities

    prove the ff identities

    sec x + 1 / tan x - tan x / sec x -1 = 0



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



    thanks
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  2. #2
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    Hi,
    Quote Originally Posted by eepyej View Post
    prove the ff identities

    sec x + 1 / tan x - tan x / sec x -1 = 0
    \frac{\sec(x)+1}{\tan(x)}-\frac{\tan(x)}{\sec(x)-1}=\frac{(\sec(x)+1) (\sec(x)-1)-\tan^2(x)}{\tan(x)(\sec(x)-1)}

    Now, expand in the numerator (\sec(x)+1)(\sec(x)-1) thanks to this formula : (a+b)(a-b)=a^2-b^2
    And remember that 1+\tan^2(x)=\sec^2(x)


    1-sinx/1+sinx = (secx-tanx)^2
    Multiply by 1=\frac{1-\sin(x)}{1-\sin(x)}
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  3. #3
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    Hello, eepyej!

    Another approach . . .


    Prove: . \frac{\sec x + 1}{\tan x} - \frac{\tan x}{\sec x -1} \;=\; 0

    Multiply the first fraction by \frac{\sec x - 1}{\sec x - 1}

    . . {\color{blue}\frac{\sec x - 1}{\sec x - 1}}\cdot\frac{\sec x + 1}{\tan x} - \frac{\tan x}{\sec x - 1} \;\;=\;\;\frac{\sec^2\!x-1}{\tan x(\sec x-1)} - \frac{\tan x}{\sec x - 1}

    . . = \;\frac{\tan^2\!x}{\tan x(\sec x - 1)} - \frac{\tan x}{\sec x - 1} \;\;=\;\;\frac{\tan x}{\sec x - 1} - \frac{\tan x}{\sec x - 1} \;\;=\;\;0

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