# Solve the given problem involving: ##\tan^{-1} (2x+1)+ \tan^{-1} (2x-1)##

• chwala
chwala
Gold Member
Homework Statement
See attached.
Relevant Equations
Trigonometry

I let

##\tan θ = 2x+1## and ##\tan β = 2x-1##

##θ + β = \tan^{-1} \left[\dfrac{(2x+1)+(2x-1)}{1- (2x+1)(2x-1)}\right]##

...

##θ + β = \tan^{-1} \left[\dfrac{4x}{1- 2x^2+1}\right]##

##θ + β = \tan^{-1} \left[\dfrac{4x}{2(1-x^2)}\right]##

then

##\tan^{-1} \left[\dfrac{4x}{2(1-x^2)}\right]= \tan^{-1}((2)##

##\dfrac{2x}{1-x^2}= 2##

##x^2+x-2=0## We shall then have ##x=-2## and ##x=1##.

##x## can only be equal to ##1##.

Check if correct fine...alternative approach/positive criticism allowed. Cheers!

chwala said:
##x## can only be equal to ##1##.
Did you verify that your proposed "solution" ##x=1## satisfies the original equation ##\tan^{-1}\left(2x+1\right)+\tan^{-1}\left(2x-1\right)=\tan^{-1}\left(2\right)##?
Hint: ##\tan^{-1}\left(3\right)+\tan^{-1}\left(1\right)=2.034## whereas ##\tan^{-1}\left(2\right)=1.107##

chwala
renormalize said:
Did you verify that your proposed "solution" ##x=1## satisfies the original equation ##\tan^{-1}\left(2x+1\right)+\tan^{-1}\left(2x-1\right)=\tan^{-1}\left(2\right)##?
Hint: ##\tan^{-1}\left(3\right)+\tan^{-1}\left(1\right)=2.034## whereas ##\tan^{-1}\left(2\right)=1.107##
I will check this as ##x## cannot be equal to ##1##. Back after break..

just amended...

I let

##\tan θ = 2x+1## and ##\tan β = 2x-1##

##θ + β = \tan^{-1} \left[\dfrac{(2x+1)+(2x-1)}{1- (2x+1)(2x-1)}\right]##

...

##θ + β = \tan^{-1} \left[\dfrac{4x}{1- 4x^2-1}\right]##

##θ + β = \tan^{-1} \left[\dfrac{4x}{4x^2}\right]##

then

##\tan^{-1} \left[\dfrac{4x}{4x^2}\right]= \tan^{-1}(2)##

##⇒\dfrac{1}{x}= 2##

##⇒x=\dfrac{1}{2}##.

Cheers!

Structure seeker
The answer is correct but the derivation isn't. Just check your evaluation of ##1-(2x+1)(2x-1)##.

Hint: there is also a negative solution

e_jane
Structure seeker said:
The answer is correct but the derivation isn't. Just check your evaluation of ##1-(2x+1)(2x-1)##.
Arrrrrgh ...I missed the negative... silly me! too much on my desk... I'll check and amend later. Cheers mate...

Structure seeker said:
The answer is correct but the derivation isn't. Just check your evaluation of ##1-(2x+1)(2x-1)##.
...
We shall have,

##8x^2+4x-4=0##

##(2x-1)(x+1)=0##

##x=\dfrac{1}{2}## and ##x=-1##.

Therefore

##x=\dfrac{1}{2}## only.

Thanks @Structure seeker ...guess my mind was off today!

Last edited:
Structure seeker said:
Hint: there is also a negative solution
Not true for this problem as worded. It starts with: "Show that there is a positive value of x..."

e_jane
chwala said:
##⇒x=\dfrac{1}{2}##.

Cheers!
chwala, you don't need to go through all the effort of deriving and factoring a quadratic equation to solve the problem:$$\tan^{-1}\left(2x+1\right)+\tan^{-1}\left(2x-1\right)=\tan^{-1}\left(2\right)$$for positive ##x##. Define ##z\equiv 2x-1## and substitute to get:$$\tan^{-1}\left(z+2\right)+\tan^{-1}\left(z\right)=\tan^{-1}\left(2\right)$$Since ##\tan^{-1}\left(0\right)=0##, by simple inspection this equation is satisfied by ##z=0\Rightarrow x=\frac{1}{2}##.

chwala

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