`f(x) = (1 + cos(x))/(1 - cos(x)), ((pi/2),1)` Find an equation of the tangent line to the graph of `f` at the given point.
`f(x) = (1+cos(x))/(1-cos(x)).`
First, let's check that `f(pi/2) = 1` : `cos(pi/2) = 0` and (1+0)/(1-0) = 1.
Then remember that
`1+cos(x) = 2*[cos(x/2)]^2` and `1-cos(x) = 2*[sin(x/2)]^2` .
Therefore `f(x) = [cot(x/2)]^2` and
`f'(x) = 2*cot(x/2)*(1/2)*(-1)/[sin(x/2)]^2 = -(cos(x/2))/(sin(x/2))^3` .
And `f'(pi/2) = -(1/sqrt(2))/(1/sqrt(2))^3 = -2.`
The tangent line goes though the point of tangency and has the slope equal to the derivative in that point. The equation is
`y - 1 = (-2)*(x - pi/2)` , or
`y = -2x +pi + 1` .