Find an equation of the tangent line to the curve?

sin(x)/(1+cosx)              P=((pi/2), 1) and the point (𝟎,𝟎).

1 Answer

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  • u = sin(x)

    u' = cos(x)

    v = 1 + cos(x)

    v' = -sin(x)

    (v * u' - u * v') / v^2 =>

    ((1 + cos(x)) * cos(x) - sin(x) * (-sin(x))) / (1 + cos(x))^2 =>

    (cos(x) + cos(x)^2 + sin(x)^2) / (1 + cos(x))^2 =>

    (cos(x) + 1) / (1 + cos(x))^2 =>

    1/(1 + cos(x))

    x = pi/2

    1/(1 + cos(pi/2)) =>

    1/(1 + 0) =>

    1/1 =>

    1

    You need a line with a slope of 1 that passes through (pi/2 , 1)

    y - 1 = 1 * (x - pi/2)

    y = x + 1 - (pi/2)

    x = 0

    1/(1 + cos(0)) =>

    1/(1 + 1) =>

    1/2

    y - 0 = (1/2) * (x - 0)

    y = x/2

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