Verify the identity: $\tan^{-1} x +\tan^{-1} (1/x) = \pi /2$

trigonometry

Verify the identity: $\tan^{-1} x + \tan^{-1} (1/x) = \frac\pi 2, x > 0$

$$\alpha= \tan^{-1} x$$

$$\beta = \tan^{-1} (1/x)$$

$$\tan \alpha = x$$

$$\tan \beta = 1/x$$

$$\tan^{-1}[\tan(\alpha + \beta)]$$

$$\tan^{-1}\left [{\tan\alpha + \tan\beta\over 1 - \tan\alpha \tan\beta} \right]$$

$$\tan^{-1}\left[ {x + 1/x\over 1- x/x }\right]$$

$$\tan^{-1}\left[{x + (1/x)\over 0} \right]$$

I can't find out what I'm doing wrong..

Solution 1:

Hint: When you want to prove that something smooth is constant, use derivatives.

details: if $f(x) = \arctan x + \arctan\frac 1x$ then $$ f'(x) = \frac 1{1+x^2} + \frac 1{1+\left(\frac 1x\right)^2}\times \left(-\frac{1}{x^2}\right) =0 $$ then $f(x) = f(1) = 2\arctan 1 = \frac\pi 2$ on the interval $\{x>0\}$.

The problem of your method is that the formula you are using is true only when $$ \alpha , \beta, \alpha + \beta \neq \frac\pi 2 \mod \pi $$

Solution 2:

An easy, mostly graphical proof: $\tan\alpha=x$, $\tan\beta=\frac1x$, and $\alpha+\beta=\frac\pi2$.

Triangle

The reason you get a division by zero in the argument of arctan is that $\displaystyle\lim_{\varphi\to\frac\pi2}\tan\varphi=\pm\infty\approx\tfrac10$. So, in very informal notation, you could say that $\tan^{-1}(\infty)=\tfrac\pi2$, and that your calculation in a way make sense.

Solution 3:

You're basically trying to compute $\tan(\pi/2)$, which doesn't exist.

If you set $\beta=\arctan(1/x)$, then $\tan\beta=1/x$, that is $$ x=\cot\beta=\tan\left(\frac{\pi}{2}-\beta\right) $$ Therefore $$ \arctan x=\arctan\tan\left(\frac{\pi}{2}-\beta\right)=\frac{\pi}{2}-\beta $$ by the hypothesis that $x>0$, so that $0<\arctan(1/x)<\pi/2$.