Find the equation of the tangent to the curve y = (sec4 x – tan4 x

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

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

Answer»

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