Question

Find the equation of the tangent to the curve y = (sec⁴ x – tan⁴ x) at x = π/3.

Answer 1

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