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Math Help - Trig Identities in longer way

  1. #1
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    Trig Identities in longer way

    I have this trig problem:

     \frac {secx - 1}{Tan x } = \frac {Tan x}{Sec x - 1}

    now what we did in class is cross multuplying both of them and we get the answer.

    I want to know how to get the answer in long process. thank you!
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  2. #2
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    Hello Anemori
    Quote Originally Posted by Anemori View Post
    I have this trig problem:

     \frac {secx - 1}{Tan x } = \frac {Tan x}{Sec x - 1}

    now what we did in class is cross multuplying both of them and we get the answer.

    I want to know how to get the answer in long process. thank you!
    I think you have a sign wrong. One of the minus signs should be a plus sign. So perhaps you must mean: prove the identity:
     \frac {\sec x - 1}{\tan x } = \frac {\tan x}{\sec x + 1}
    Start with the LHS, and multiply top-and-bottom of the fraction by (\sec x + 1)
     \frac {\sec x - 1}{\tan x } =\frac {(\sec x - 1)\color{red}(\sec x + 1)}{\tan x\color{red} (\sec x + 1)}
    =\frac{\sec^2x-1}{\tan x (\sec x + 1)}

    =\frac{\tan^2x}{\tan x (\sec x + 1)}, using the identity \sec^2x -1 = \tan^2x


    =\frac{\tan x}{\sec x + 1}

    On the other hand, if it's the other minus sign that's wrong, again you'll start with the LHS, but this time multiply top-and-bottom by (\sec x - 1).

    Grandad
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  3. #3
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    Quote Originally Posted by Grandad View Post
    Hello AnemoriI think you have a sign wrong. One of the minus signs should be a plus sign. So perhaps you must mean: prove the identity:
     \frac {\sec x - 1}{\tan x } = \frac {\tan x}{\sec x + 1}
    Start with the LHS, and multiply top-and-bottom of the fraction by (\sec x + 1)
     \frac {\sec x - 1}{\tan x } =\frac {(\sec x - 1)\color{red}(\sec x + 1)}{\tan x\color{red} (\sec x + 1)}
    =\frac{\sec^2x-1}{\tan x (\sec x + 1)}

    =\frac{\tan^2x}{\tan x (\sec x + 1)}, using the identity \sec^2x -1 = \tan^2x


    =\frac{\tan x}{\sec x + 1}

    On the other hand, if it's the other minus sign that's wrong, again you'll start with the LHS, but this time multiply top-and-bottom by (\sec x - 1).

    Grandad
    You were right with the sign. btw what is LHS and why multiplying by the denominator of the other side?
    Last edited by Anemori; March 16th 2010 at 09:40 PM.
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