Consider y = cot2 x – 2 cot x + 2 as the equation of curve
Take the value of x = π/4 so we get y = 1
By differentiating both sides w.r.t. x
dy/dx = 2 cot x(- cosec2 x) – 2(- cosec2 x)
By further simplification
dy/dx = – 2 cot x cosec2 x + 2 cosec2 x
We get
(dy/dx)(x = π/4) = 0
Here the equation of tangent at point (π/4, 1)
y – 1 = 0(x – π/4)
So we get
y – 1 = 0
Here y = 1
Here the equation of normal at point (π/4, 1)
y – π/4 = 1/0(x – π/4)
We get
0 = x – π/4
Here x = π/4