Question

What is the equation of the line normal to `f(x)=secx-cot^2x` at `x=pi/3`?

Answer 1

The equation of normal is `3 x +15 y = 28.14`

Explanation:

`sec (pi/3)= 2 , tan (pi/3)= sqrt3 , cot (pi/3)= 1/sqrt3`

`cot^2 (pi/3)= 1/3 ,csc (pi/3)= 2/sqrt3, csc^2 (pi/3)= 4/3,`

`f(x) = sec x- cot ^2 x :. f(pi/4) = sec (pi/4)- cot ^2 (pi/4)` or

`f(pi/4) = 2-1/3 =5/3` The point at which normal to be drawn

is `(pi/3 , 5/3) ; f(x)=sec x- cot ^2 x`

`f'(x)=sec x tan x - 2 cot x * (-csc^2x)` or

`f'(x)=sec x tan x + 2 cot x * csc^2 x`

`f'(pi/4)=sec (pi/4) tan (pi/4) + 2 cot (pi/4) * csc^2 (pi/4)`or

`f'(pi/4)=2sqrt3 + 2/sqrt3 * 4/3 = 5.00` The slope of

tangent at `pi/3` is `5.0` therefore, slope of normal at `pi/3`

is `(-1/5)` The equation of normal to `f(x)` at point

`(pi/3 , 5/3)` is `y- 5/3 = -1/5 (x- pi/3)` or

`15 y- 25 = -3 x + pi or 3 x +15 y = 25 +pi` or

`3 x +15 y = 28.14`

The equation of normal is `3 x +15 y = 28.14` [Ans]