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Math Help - Trigonometry problem

  1. #1
    Newbie Adikz's Avatar
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    Trigonometry problem

    Can someone help me to simplify this?

    \frac{\text{1 + secA}}{\text{sinA+tanA}}
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  2. #2
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    Re: Trigonometry problem

    Quote Originally Posted by Adikz View Post
    Can someone help me to simplify this?

    \frac{\text{1 + secA}}{\text{sinA+tanA}}
    There are MANY equivalent expressions, any of which could be considered "simpler"...

    \displaystyle \begin{align*} \frac{1 + \sec{A}}{\sin{A} + \tan{A}} &\equiv \frac{1 + \frac{1}{\cos{A}}}{\sin{A} + \frac{\sin{A}}{\cos{A}}} \\ &\equiv \frac{\frac{\cos{A} + 1}{\cos{A}}}{\frac{\sin{A}\cos{A} + \sin{A}}{\cos{A}}} \\ &\equiv \frac{\cos{A} + 1}{\sin{A}\cos{A} + \sin{A}} \\ &\equiv \frac{\cos{A} + 1}{\sin{A}\left( \cos{A} + 1\right)} \\ &\equiv \frac{1}{\sin{A}} \\ &\equiv \csc{A} \end{align*}
    Thanks from Adikz
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  3. #3
    Newbie Adikz's Avatar
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    Re: Trigonometry problem

    Thanks.. Can I ask for more?
    -secAcscA ( 1-2sin^2A )
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  4. #4
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    Re: Trigonometry problem

    -\sec{x}\csc{x}(1-2\sin^2{x})

    -\frac{1}{\cos{x}} \cdot \frac{1}{\sin{x}} \cdot \cos(2x)

    -\frac{2\cos(2x)}{2\cos{x}\sin{x}}

    -\frac{2\cos(2x)}{\sin(2x)}

    -2\cot(2x)
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