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Math Help - Prove tan

  1. #1
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    Prove tan

    Prove that tan(\theta+45degrees) = \frac{1+tan\theta}{1-tan\theta}
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  2. #2
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    straightforward, just use the formula for \tan(x+y).
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  3. #3
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    I've never done this type question before, could you post how its done please. I know I have to use the formula but how is it done?
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  4. #4
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    \tan(x+y)=\frac{\tan x+\tan y}{1-\tan(x)\tan(y)}.
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  5. #5
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    So what is x and y? Is it 1+tan\theta and 1-tan\theta?
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  6. #6
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    Quote Originally Posted by gary223 View Post
    So what is x and y? Is it 1+tan\theta and 1-tan\theta?
    \tan(\theta + 45^\circ) = \frac{\tan{\theta} + \tan(45^\circ)}{1 - \tan{\theta} \cdot \tan(45^\circ)}
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  7. #7
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    Quote Originally Posted by skeeter View Post
    \tan(\theta + 45^\circ) = \frac{\tan{\theta} + \tan(45^\circ)}{1 - \tan{\theta} \cdot \tan(45^\circ)}
    Okay, I see. So whats the next step? Do I cancel some parts out or do I calculate each part? Or is that the question finished?
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  8. #8
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    Quote Originally Posted by gary223 View Post
    Okay, I see. So whats the next step? Do I cancel some parts out or do I calculate each part? Or is that the question finished?
    substitute in the value of \tan(45^\circ).
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  9. #9
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    Quote Originally Posted by skeeter View Post
    \tan(\theta + 45^\circ) = \frac{\tan{\theta} + \tan(45^\circ)}{1 - \tan{\theta} \cdot \tan(45^\circ)}
    So its \tan(\theta + 45^\circ)= \frac{tan\theta+1}{1-tan\theta.1}

    Correct?
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  10. #10
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    Quote Originally Posted by gary223 View Post
    So its \tan(\theta + 45^\circ)= \frac{tan\theta+1}{1-tan\theta.1}

    Correct?
    look again at what you were trying to prove ...
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  11. #11
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    Quote Originally Posted by skeeter View Post
    look again at what you were trying to prove ...
    \theta + 1= \frac{tan\theta+1}{1-tan\theta.1}

    ???

    EDIT:

    Or is it:

    tan\theta + tan1= \frac{tan\theta+1}{1-tan\theta.1}
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  12. #12
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    Or is it tan\theta+ 1 = \frac{tan\theta+1}{1-tan\theta.1}

    Sorry for double post but I wanna get this.
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  13. #13
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    you were asked to prove that \tan(\theta+45^\circ) = \frac{1+\tan{\theta}}{1-\tan{\theta}}

    Krizalid gave you the sum identity for tangent ...

    \tan(x+y) = \frac{\tan{x} + \tan{y}}{1 - \tan{x} \cdot \tan{y}}<br />

    my initial post basically told you to substitute \theta for x and 45^\circ for y ...

    \tan(\theta+45^\circ) = \frac{\tan{\theta} + \tan(45^\circ)}{1 - \tan{\theta} \cdot \tan(45^\circ)}

    using the fact that \tan(45^\circ) = 1 ...

    \tan(\theta+45^\circ) = \frac{\tan{\theta} + 1}{1 - \tan{\theta}} = \frac{1+\tan{\theta}}{1 - \tan{\theta}}

    the original equation is proved ... you're done.
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  14. #14
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    Wow, it seems so easy once you do it. Thanks.
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  15. #15
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    Note that special angles exist, which are basically, sine, tangent and cosine of 30, 45, 60and 90. And you are required to recognise them..
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