It is given that y = (sec4 x – tan4 x)
By substituting the value of x = π/3
y = (sec4 (π/3) – tan4 (π/3)) = 24 – (√3)4 = 7
By differentiating both sides w.r.t x
dy/dx = 4 sec3 x sec x tan x – 4 tan3 x sec2 x
It can be written as
dy/dx = 4 sec2 x tan x (sec2 x – tan2 x) = 4 sec2 x tan x
So (dy/dx)x= π/3 = 4 sec2 (π/3) tan (π/3) = 16 √3
Here the equation of tangent at point (π/3, 7)
y – 7 = 16√3(x – π/3)
On further calculation
y – 7 = 16√3x – 16√3 π/3
By taking LCM as 3
y – 7 = (48√3x – 16√3 π)/3
We get
3y – 21 = 48√3x – 16√3 π
We can write it as
48√3x – 3y – 16√3 π + 21 = 0
Taking negative sign as common
3y – 48√3x + 16√3 π – 21 = 0