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प्रश्न
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)`.
उत्तर १
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`
उत्तर २
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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