Question

What is the equation of the line normal to `f(x)=(xsinx)/tanx` at `x=pi/3`?

Answer 1

Equation of normal is `(y-pi/6)=(2sqrt3)/(pi-sqrt3)xx(x-pi/3)`

Explanation:

As `f(x)=(xsinx)/tanx`, at `x=pi/3`, `f(x)=(pi/3xxsqrt3/2)/sqrt3=pi/6`

Hence,m we are seeking equation of normal at `(pi/3,pi/6)`

As slope of tangent is given by `f'(x)`, as `f(x)=(xsinx)/(sinx/cosx)=xcosx`

Hence `f'(x)=cosx-xsinx` and at `x=pi/3`, `f'(pi/3)=cos(pi/3)-pi/3xxsin(pi/3)=1/2-pi/3xxsqrt3/2`

= `1/2-pi/(2sqrt3)=(sqrt3-pi)/(2sqrt3)`

Hence slope of tangent is `(sqrt3-pi)/(2sqrt3)`

and that of normal would be `(-1)/((sqrt3-pi)/(2sqrt3))=(2sqrt3)/(pi-sqrt3)`

and equation of normal would be

`(y-pi/6)=(2sqrt3)/(pi-sqrt3)xx(x-pi/3)`