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Math Help - Proving identitiees

  1. #1
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    Proving identitiees

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

    x there is θ. I don't know how to put the θ.

    Thank you.!
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  2. #2
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    Quote Originally Posted by reiward View Post
    \frac{sec x + 1}{tan x} = \frac{tan x}{sec x -1}

    x there is θ. I don't know how to put the θ.

    Thank you.!
    Substitute \tan x = \frac{\sin x}{\cos x} and \sec x = \frac{1}{\cos x} and simplify:

    LHS = \frac{1 + \cos x}{\sin x}

    RHS = \frac{\sin x}{1 - \cos x} = \frac{\sin x}{(1 - \cos x)} \cdot \frac{(1 + \cos x)}{(1 + \cos x)} = \frac{\sin x (1 + \cos x)}{(1 - \cos^2 x)} and it should be clear how to continue and end up with the LHS.
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    Quote Originally Posted by mr fantastic View Post
    Substitute \tan x = \frac{\sin x}{\cos x} and \sec x = \frac{1}{\cos x} and simplify:

    LHS = \frac{1 + \cos x}{\sin x}

    RHS = \frac{\sin x}{1 - \cos x} = \frac{\sin x}{(1 - \cos x)} \cdot \frac{(1 + \cos x)}{(1 + \cos x)} = \frac{\sin x (1 + \cos x)}{(1 - \cos^2 x)} and it should be clear how to continue and end up with the LHS.
    Pardon me, but what is LHS and RHS?

    OK, so I get the process of ending up with the LHS you were saying. But in our class, we need to retain the original term of the left or right side. So I need to show that \frac{sec x + 1}{tan x} = \frac{sec x + 1}{tan x}<br />

    So from the LHS, can you show me how to make the LHS into \frac{sec x + 1}{tan x}
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  4. #4
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    Quote Originally Posted by reiward View Post
    Pardon me, but what is LHS and RHS?

    OK, so I get the process of ending up with the LHS you were saying. But in our class, we need to retain the original term of the left or right side. So I need to show that \frac{sec x + 1}{tan x} = \frac{sec x + 1}{tan x}<br />

    So from the LHS, can you show me how to make the LHS into \frac{sec x + 1}{tan x}
    Divide numerator and denominator by \cos x.
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