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

[chwala]

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!

 

[renormalize]

$x$ 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]

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..

 

[chwala]

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 $1-(2x+1)(2x-1)$.

 

[Structure seeker]

Hint: there is also a negative solution

 

[chwala]

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...

 

[chwala]

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!

 

[Structure seeker]

Although I'm sure you know the math needed already very well, there's a small mistake in here still. But speaking of them, here's a nice story:
https://www.restorationadvisers.com/blog/einsteinlesson

 

[renormalize]

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..."

 

[renormalize]

$⇒x=\dfrac{1}{2}$.

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 $z=0\Rightarrow x=\frac{1}{2}$.