Consider y = (sin 2x + cot x + 2)2 as the equation of the curve
By substituting the value of x = π/2 we get y = 4
On differentiation of both sides w.r.t. x
dy/dx = 2(sin 2x + cot x + 2) (2 cos 2x – cosec2 x)
We get
(dy/dx)x= π/2 = – 12
Here the equation of normal at point (π/2, 4)
y – 4 = 1/12 (x – π/2)
On further calculation
12y – 48 = x – π/2
We get
24y – 96 = 2x – π
It can be written as
24y – 96 – 2x + π = 0
24y – 2x + π – 96 = 0