# If Tan-1 X - Cot-1 X = Tan-1 ( 1 √ 3 ) , X> 0 Then Find the Value of X and Hence Find the Value of Sec-1 ( 2 X ) - Mathematics

#### Question

Sum

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).

#### Solution 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

#### Solution 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

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