Question

If tan^-1 x - cot^-1 x = tan^-1 `(1/sqrt(3)),`x> 0 then find the value of x and hence find the value of sec^-1 `(2/x)`.

Answer 1

tan^-1 x - cot^-1 x = tan^-1 `(1/sqrt(3)),`x> 0

⇒ tan^-1 x - tan^-1 `(1/"x")` = tan^-1 `(1/sqrt(3))`   ....[∵ cot^-1 "x" = tan^-1 `(1/"x"), "x" >0`] 

⇒`tan^-1 (("x"-1/"x")/(1+"x". 1/"x")) = tan^-1 (1/sqrt3)`

⇒ `("x"^2 - 1)/(2"x") = 1/sqrt(3)`

⇒ `sqrt3"x"^2 - 2"x" - sqrt(3) = 0`

⇒ `sqrt3"x"^2 - 3"x" + "x" -sqrt(3) = 0`

⇒ `sqrt3x ("x" -sqrt3) + 1 ("x" - sqrt3) = 0`

⇒`(x - sqrt3) (sqrt3"x" + 1 ) =0`

⇒ `"x" = - 1/sqrt3, sqrt3`

∵ x >0, x = `sqrt3`

⇒ `sec^-1 (2/"x") = sec^-1 (2/sqrt3)`

⇒ `sec^-1 (2/"x") = sec^-1 (sec  π/(6))` 

⇒ `sec^-1 (2/"x") = π/6`

Answer 2

Given,

tan^-1 x - cot^-1 x = tan^-1 `(1/sqrt3),` x > 0

⇒ `tan^-1 x - tan^-1 (1/x) = tan^-1 (1/sqrt3)   ....[ ∵ cot^-1 x = tan-1 (1/x), x > 0 ] `

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

⇒ `("x"^2 - 1)/(2"x") = 1/sqrt(3)`

⇒ `sqrt3"x"^2 - 2"x" - sqrt(3) = 0`

⇒ `sqrt3"x"^2 - 3"x" + "x" -sqrt(3) = 0`

⇒ `sqrt3x ("x" -sqrt3) + 1 ("x" - sqrt3) = 0`

⇒`(x - sqrt3) (sqrt3"x" + 1 ) =0`

⇒ `"x" = - 1/sqrt3, sqrt3`

∵ x >0, x = `sqrt3`

⇒ `sec^-1 (2/"x") = sec^-1 (2/sqrt3)`

⇒ `sec^-1 (2/"x") = sec^-1 (sec  π/(6))` 

⇒ `sec^-1 (2/"x") = π/6`