Question

Solve the following equation for x:

 cot⁻¹x − cot⁻¹(x + 2) =`pi/12`, x > 0

Answer 1

⇒ `cot^-1(x)-cot^-1(x+2)=pi/12`

⇒ `tan^-1(1/x)+cot^-1(1/(x+2))=pi/12`     `[because cot^-1x=tan^-1 1/x]`

⇒ `tan^-1((1/x-1/(x+2))/(1+1/(x(x+2))))=pi/12`

⇒ `tan^-1((2/(x(x+2)))/((x^2+2x+1)/(x(x+2))))=pi/12`

⇒ `tan^-1(2/(x^2+2x+1))=pi/12`

⇒ `(2/(x^2+2x+1))=tan  pi/12`

⇒ `(2/(x^2+2x+1))=tan(pi/3-pi/4)`

⇒ `(2/(x^2+2x+1))=(tan  pi/3-tan  pi/4)/(1+tan  pi/3xxtan  pi/4`

⇒ `(2/(x^2+2x+1))=(sqrt3-1)/(sqrt3+1)`

⇒ `(2/(x^2+2x+1))=(sqrt3-1)/(sqrt3+1)xx(sqrt3+1)/(sqrt3+1)`

⇒ `(2/(x^2+2x+1))=2/(sqrt3+1)^2`

⇒ `1/(x+1)^2=1/(sqrt3+1)^2`

⇒ `x+1=sqrt3+1`

⇒ `x=sqrt3`