How to prove $\tan^{-1}(n+1)-\tan^{-1}(n-1)=\tan^{-1}\big(\frac{2}{n^2}\big)$?

Question

Prove $$\tan^{-1}(n+1)-\tan^{-1}(n-1)=\tan^{-1}\big(\frac{2}{n^2}\big)$$ for $n \ge 1$

If I use mathematical induction how do I manipulate the numbers to fit in the induction hypothesis?

Is there another way?

Answer 1

Here are two methods to do it.

Method #1.

Using the identity:

$$\tan (\alpha - \beta) = \frac{\tan \alpha - \tan \beta}{1 + \tan \alpha \tan \beta}$$

Then:

$$\frac{ (n+1) - (n-1) } {1 + (n+1)(n-1)} = \frac{2}{n^2}$$ $$\Rightarrow \tan\left( \arctan (n+1) - \arctan (n-1) \right) = \frac{2}{n^2}$$ $$\Rightarrow \arctan (n+1) - \arctan (n-1) = \arctan \frac{2}{n^2}$$

Method #2.

Using the identity:

$$\arctan x = \frac{i}{2} \log \left( \frac{1 - ix}{1 + ix} \right)$$

where $i^2 = -1$. Then:

$$\arctan (n+1) - \arctan (n-1)$$ $$ = \frac{i}{2} \log \left( \frac{1 - i(n+1)}{1 + i(n+1)} \right) - \frac{i}{2} \log \left( \frac{1 - i(n-1)}{1 + i(n-1)} \right) $$ $$ = \frac{i}{2} \log \left( \frac{1 - i(n+1)}{1 + i(n+1)} \frac{1 + i(n-1)}{1 - i(n-1)} \right) $$ $$ = \frac{i}{2} \log \left( \frac{n^2 - 2i}{n^2 + 2i} \right)$$ $$ = \frac{i}{2} \log \left( \frac{1 - i \frac{2}{n^2}}{1 + i \frac{2}{n^2}} \right) $$ $$ = \arctan \frac{2}{n^2} $$

Answer 2

We have$\def\LHS{{\rm LHS}}\def\RHS{{\rm RHS}}$ $$\eqalign{\tan(\LHS) &=\frac{\tan(\tan^{-1}(n+1))-\tan(\tan^{-1}(n-1))}{1+\tan(\tan^{-1}(n+1))\tan(\tan^{-1}(n-1))}\cr &=\frac{(n+1)-(n-1)}{1+(n+1)(n-1)}\cr &=\frac{2}{n^2}\cr &=\tan(\RHS)\ .\cr}$$ Also, $$0\le\tan^{-1}(n\pm1)<\frac{\pi}{2}\ ,$$ so $$-\frac{\pi}{2}<\LHS<\frac{\pi}{2}\ ,$$ and $$-\frac{\pi}{2}<\RHS<\frac{\pi}{2}\ .$$ Therefore $\tan(\LHS)=\tan(\RHS)$ implies that $\LHS=\RHS$.

Answer 3

Hint: $$\tan(\tan^{-1}(n+1)-\tan^{-1}(n-1))=\tan\left(\tan^{-1}\left(\frac{2}{n^2}\right)\right)$$ then use the fact that $$\tan(\tan^{-1}(x))=x$$ and $$\tan(a-b)=\frac{\tan a-\tan b}{1+\tan a\tan b}$$

Answer 4

In fact $$\tan^{-1} a +\tan^{-1} b =\tan^{-1}\left (\frac{a+b}{1-ab}\right)$$
This plus $-\tan^{-1} a =\tan^{-1} (-a)$ gives your result immediately.

Answer 5

on the $y$-axis let the points $P,Q,R$ have coordinates $(0,n-1), (0,n)$ and $(0,n+1)$ respectively, and on the $x$-axis let the point $(1,0)$ be denoted A. then $$ \tan O \hat A P = n -1 \\ \tan O \hat A R = n +1 $$ so we need to show that: $$ \tan P \hat A R = \frac2{n^2} $$ a little algebra establishes that $Z$, the center of the circle through $A,P$ and $R$ has coordinates $(\frac{n^2}2, n)$

now by a theorem of elementary geometry the angle $P \hat AR$ is equal to half the angle $P \hat ZR$, which makes it true that: $$ P \hat AR = Q \hat Z P $$ and from the right-angled triangle $PQZ$ we have $\tan Q \hat ZR = \frac2{n^2}$

Answer 6

Proof without words (except these):

$$\tan\alpha = n+1 \qquad \tan\beta = n-1$$ $$\tan(\alpha-\beta) = \frac{|\overline{AD}|}{|\overline{OD}|}= \frac{|\overline{AD}||\overline{OB}|}{|\overline{OD}||\overline{OB}|}=\frac{2\cos\beta\cdot\sec\beta}{|\overline{OT}|^2}=\frac{2}{n^2}$$