Why Does This Inverse Trigonometry Problem Have Four Solutions?

[vkash]

Homework Statement

tan^-1[2x/(1-x²)]+cot^-1[(1-x²)/2x]=2π/3

2. The attempt at a solution


tan^-1[2x/(1-x²)]+cot^-1[(1-x²)/2x]=2π/3
tan^-1[2x/(1-x²)]=2π/6
take tan on both sides
2x/(1-x²) =sqrt(3)
quadratic equation so it should have 2 solutions(sqrt(3) and sqrt(1/3)).But this question has four solution.. Where am i missing solutions. Can you please help me to figure out the error..
thanks
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vikash

 

[eumyang]

According to Wolframalpha, there are two solutions, 1/√3 and -√3. How do you know that there are four solutions?

 

[Bread18]

tan^-1[2x/(1-x²)]+cot^-1[(1-x²)/2x]=2π/3
tan^-1[2x/(1-x²)]=2π/6

How did you get from the first line to the second?

 

[SammyS]

How did you get from the first line to the second?

cot^-1(a/b) = tan^-1(b/a)

I.e.: cot(x) = 1/tan(x)

 

[vkash]

According to Wolframalpha, there are two solutions, 1/√3 and -√3. How do you know that there are four solutions?

Thanks for link.
There are more two answers in book that's why i ask. But now they will be marked as wrong.

 

[Bread18]

cot^-1(a/b) = tan^-1(b/a)

I.e.: cot(x) = 1/tan(x)

Ah right, missed that. Thanks.