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Math Help - Not Sure if I did this correctly with Trig Identities

  1. #1
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    Not Sure if I did this correctly with Trig Identities

    Algebra seems like a blur, so I need a bit of help with this problem.(~)==theta
    The original problem is this, and you are supposed to prove that it is true

    cos(~)
    ------- = Sec(~)+tan(~)
    1-sin(~)

    my work:

    1/sec * 1/1-sin(~)

    1/sec(1-sin(~)

    1/sec(~)-sec(~)*sin(~)

    ----This is where I got confused(I couldn't remember if you could use the inverse to make this statement true)------

    (1/sec(~)-tan(~))^-1=Sec(~)+tan(~)

    Could I get a yes or no if I did it right? Thanks
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  2. #2
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    Re: Not Sure if I did this correctly with Trig Identities

    \frac{\cos{t}}{1 - \sin{t}} \cdot \frac{1+\sin{t}}{1+\sin{t}} =

    \frac{\cos{t}(1+\sin{t})}{1-\sin^2{t}} =

    \frac{\cos{t}(1+\sin{t})}{\cos^2{t}} =

    \frac{1+\sin{t}}{\cos{t}} =

    \frac{1}{\cos{t}} + \frac{\sin{t}}{\cos{t}} =

    \sec{t} + \tan{t}
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  3. #3
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    Re: Not Sure if I did this correctly with Trig Identities

    Quote Originally Posted by kickyou77 View Post
    Algebra seems like a blur, so I need a bit of help with this problem.(~)==theta
    The original problem is this, and you are supposed to prove that it is true

    cos(~)
    ------- = Sec(~)+tan(~)
    1-sin(~)

    my work:

    1/sec * 1/1-sin(~)

    1/sec(1-sin(~)

    1/sec(~)-sec(~)*sin(~)

    ----This is where I got confused(I couldn't remember if you could use the inverse to make this statement true)------

    (1/sec(~)-tan(~))^-1=Sec(~)+tan(~)

    Could I get a yes or no if I did it right? Thanks
    \displaystyle \begin{align*} \frac{\cos{\theta}}{1 - \sin{\theta}} &\equiv \frac{\cos{\theta} \left( 1 + \sin{\theta} \right)}{\left( 1 - \sin{\theta} \right) \left( 1 + \sin{\theta} \right)} \\ &\equiv \frac{\cos{\theta}\left( 1 + \sin{\theta} \right)}{1 - \sin^2{\theta}} \\ &\equiv \frac{\cos{\theta}  \left( 1 + \sin{\theta} \right)}{\cos^2{\theta}} \\ &= \frac{1 + \sin{\theta}}{\cos{\theta}} \\ &\equiv \frac{1}{\cos{\theta}} + \frac{\sin{\theta}}{\cos{\theta}} \\ &\equiv \sec{\theta} + \tan{\theta} \end{align*}
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  4. #4
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    Re: Not Sure if I did this correctly with Trig Identities

    Hello, kickyou77!

    \text{Prove: }\:\frac{\cos\theta}{1-\sin\theta} \:=\:\sec\theta + \tan\theta

    You can also work from the right side . . .

    \sec\theta + \tan\theta \;=\; \frac{1}{\cos\theta} + \frac{\sin\theta}{\cos\theta} \;=\;\frac{1 + \sin\theta}{\cos\theta}

    . . . . . . . . . =\;\frac{1 + \sin\theta}{\cos\theta}\cdot {\color{blue}\frac{1-\sin\theta}{1-\sin\theta}} \;=\;\frac{1-\sin^2\!\theta}{\cos\theta(1 - \sin\theta)}

    . . . . . . . . . =\;\frac{\cos^2\!\theta}{\cos\theta(1-\sin\theta)} \;=\; \frac{\cos\theta}{1-\sin\theta}
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