Question

Solve: `cot^-1 x - cot^-1 (x + 2) = pi/12, x > 0`

Answer 1

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

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

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

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

⇒ `2/(x^2 + 2x + 1) = tan15^circ`

We know that, tan 15° = `2 - sqrt(3)`

⇒ `2/(x^2 + 2x + 1) = 2 - sqrt(3)`

⇒ `x^2 + 2x + 1 = 2/(2 - sqrt(3)`

⇒ `(x + 1)^2 = 2/(2 - sqrt(3)) xx [(2 + sqrt(3))/(2 + sqrt(3))]`

⇒ `(x + 1)^2 = (2(2 + sqrt(3)))/(4 - 3)`

⇒ (x + 1)² = `4 + 2sqrt(3)`

⇒ (x + 1)² = `1 + 3 + 2sqrt(3)`

⇒ (x + 1)² = `(1 + sqrt(3))^2`

⇒ x + 1 = `1 + sqrt(3)`

∴ x = `sqrt(3)`