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Thread: Trig identity

  1. #1
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    Trig identity

    Hi,
    Have this problem not sure if i've answered the question or not?

    Verify the identity; (secA - 1) / (secA +1) + (cosA -1) / (cos A + 1) = 0

    i changed secA into 1/cosA then re arrange and manipulated the LHS so at the end i get 1 - cos^2(x) = 1 - cos^2(x) or that the lhs had = 0 but not sure if i have properly answered the question?
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  2. #2
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    Quote Originally Posted by monster View Post
    Hi,
    Have this problem not sure if i've answered the question or not?

    Verify the identity; (secA - 1) / (secA +1) + (cosA -1) / (cos A + 1) = 0

    i changed secA into 1/cosA then re arrange and manipulated the LHS so at the end i get 1 - cos^2(x) = 1 - cos^2(x) or that the lhs had = 0 but not sure if i have properly answered the question?
    \frac{\sec{A} - 1}{\sec{A} + 1} + \frac{\cos{A} - 1}{\cos{A} + 1} = \frac{\frac{1}{\cos{A}} - 1}{\frac{1}{\cos{A}} + 1} + \frac{\cos{A} - 1}{\cos{A} + 1}

     = \frac{\frac{1 - \cos{A}}{\cos{A}}}{\frac{1 + \cos{A}}{\cos{A}}} + \frac{\cos{A} - 1}{\cos{A} + 1}

     = \frac{1 - \cos{A}}{1 + \cos{A}} + \frac{\cos{A} - 1}{\cos{A} + 1}

     = \frac{1 - \cos{A} + \cos{A} - 1}{1 + \cos{A}}

     = \frac{0}{1 + \cos{A}}

     = 0.
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