Solve the equation (2/arctanx)-arctanx=1

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justaguide | College Teacher | (Level 2) Distinguished Educator

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We have the equation 2/arc tanx - arc tanx = 1 to solve.

let y = arc tan x.

2/arc tanx - arc tanx = 1

=> 2/y - y = 1

mutiply all the terms with y

=> 2 - y^2 = y

move the terms to one side

=> y^2 + y - 2  = 0

y1 = [-b + sqrt ( b^2 - 4ac] / 2a

= [ -1 + sqrt (1 + 8)] / 2

= [ -1 + sqrt 9]/2

=> 1

y2 = -1/2 - 3 / 2

y2 = -2

As y = arc tan x

arc tan x = 1

=> x = tan 1

arc tan x = -2

=> x = - tan 2

Therefore x is equal to tan 1 and -tan 2.

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neela | High School Teacher | (Level 3) Valedictorian

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To solve 2/arc tanx -arctanx = 1.

We multiply both sides by arc tanx .

2 - (arc tanx)^2 = arc tanx

 2-y^2= y, where y = arc tanx.

=>  y^2+y-2 = 0.

=> (y-1)(y+2) = 0.

=> y -1 = o, or y+2 = 0.

=> y= 1, or y = -2.

 y = 1 gives arc tanx = 1. x = tan1 = 1.5574.

y = -2 gives arc tan x = -2. So x = tan (-2) = 2.18504.

Therefore x= 1.5574, or x =  2.18504.

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

We'll substitute the function arctan x = t and we'll re-write the equation:

2/t - t = 1

We'll multiply both sides by t:

2 - t^2 = t

We'll subtract t and we'll re-write the terms:

-t^2 - t + 2 = 0

We'll multiply by -1:

t^2 + t - 2 = 0

We'll apply the quadratic formula:

t1 = [-1 + sqrt(1 + 8)]/2

t1 = (-1+3)/2

t1 = 1

t2 = (-3-1)/2

t2 = -2

We'll put arcatn x = t1

arctan x = 1

x = tan1

arctan x = -2

x = tan (-2) = - tan 2

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