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I would like to know how this equality holds.

$$ \tan^{-1} \frac{(2n+1) - (2n-1)}{1 + (2n+1)(2n-1)} = \tan^{-1} \frac{1}{2n-1} - \tan^{-1} \frac{1}{2n+1}.$$

I was told to use the double angle formula for $\tan \theta$ but I can't seem to show this.

Thank you.

For some context, I was asked to show

$$ \sum_{n=1}^\infty \tan^{-1} \frac{1}{2n^2} = \frac{\pi}{4} $$

Jai Pancholi
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  • More or less take the $\tan$ of both sides, and use $\tan(u+v)=\frac{\tan u+\tan v}{1-\tan u\tan v}$. – André Nicolas Mar 29 '15 at 02:20
  • You are welcome. Note there is a tiny bit to do after the calculation, for $\tan^{-1}(x)$ is the angle between $-\pi/2$ and $\pi/2$ whose $\tan$ is $x$. So we have to show that the right-hand side is in this interval. Since $n$ is presumably a positive integer, this is straightforward. – André Nicolas Mar 29 '15 at 02:31

4 Answers4

2

The RHS sign is wrong, it should be:

$$ \tan^{-1} \frac{(2n+1) - (2n-1)}{1 + (2n+1)(2n-1)} = \tan^{-1} \frac{1}{2n-1} - \tan^{-1} \frac{1}{2n+1}.$$

The Rule is simply

$$ \tan^{-1} \frac{u -v}{1+u v} = \tan^{-1} u - \tan^{-1} v $$

Narasimham
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Hint: Apply the inverse function of $\tan^{-1}(x)$ which is $\tan(x)$ to both sides of the equation and then you can use the double angle formula.

science
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1

Observe that $\tan(x-y)=\frac{\tan(x)-\tan(y)}{1+\tan(x)\tan(y)}$. Then, letting $x=\arctan(\frac{1}{2n+1})$ and $y=\arctan(\frac{1}{2n-1})$, we see that

$$\begin{align} \tan(x-y)&=\frac{\tan(x)-\tan(y)}{1+\tan(x)\tan(y)}\\ &=\frac{\frac{1}{2n+1}-\frac{1}{2n-1}}{1+\frac{1}{2n+1}\frac{1}{2n-1}}\\ &=\frac{(2n-1)-(2n+1)}{1+(2n-1)(2n+1)} \end{align}$$

Taking the inverse tangent of both sides of this last expression reveals

$$\begin{align} x-y&=\arctan\left(\frac{1}{2n+1}\right)-\arctan\left(\frac{1}{2n-1}\right)\\ &=\arctan\left(\frac{(2n-1)-(2n+1)}{1+(2n-1)(2n+1)}\right) \end{align}$$

Mark Viola
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0

$$ \tan^{-1} \frac1{2n-1} = \arg (2n-1 + i) \\ \tan^{-1} \frac1{2n+1} = \arg (2n+1 + i) $$ hence $$ \tan^{-1} \frac1{2n-1} - \tan^{-1} \frac1{2n+1} = \arg \frac{2n-1+i}{2n+1+i} \\ = \arg(2n-1+i)(2n+1-i) = \arg (4n^2 +2i) =\tan^{-1}\frac1{2n^2} \\ =\tan^{-1}\frac{(2n+1)-(2n-1)}{1+(2n-1)(2n+1)} $$

David Holden
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