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Math Help - parametric equation in trigonometric

  1. #1
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    Exclamation parametric equation in trigonometric

    if x = a sin c and y = b tan c show that (a^2)/(x^2) - (b^2)/(y^2) = 1

    where x, y, a, b and c are just pronumerals...

    please answer
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  2. #2
    Senior Member DivideBy0's Avatar
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    Let's write this out fully:

    a^2 / (a*sin(c))^2 - b^2 / (b*sin(c))^2 = 1
    a^2 / (a^2*sin^2(c)) - b^2 / (b^2*sin^2(c)) = 1

    The a^2 and b^2's cancel out

    1/sin^2(c) - 1/tan^2(c) = 1
    1/sin^2(c) - cos^2(c)/sin^2(c) = 1

    Under the same denominator...

    (1 - cos^2(c)) / sin^2(c) = 1

    Since we know sin^2(c) + cos^2(c) = 1 (Pythagorean identity),
    1 - cos^2(c) = sin^2(c)

    Thus, we have :

    sin^2(c) / sin^2(c) = 1
    1=1

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  3. #3
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    Hello, T_o_M!

    If x = asinθ and y = btanθ, show that: .a/x - b/y .= .1

    x = asinθ . . cscθ = a/x .[1]

    y = btanθ . . cotθ = b/y .[2]


    Square [1]: .a/x .= .cscθ
    Square [2]: .b/y .= .cotθ

    Subtract: .a/x - b/y .= .cscθ - cotθ

    Since cscθ - cotθ = 1, we have: .a/x - b/y .= .1

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