# 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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- 2 months ago
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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