How can I show that $\arctan (x) + \arctan(1/x) =\frac{\pi}{2}$?
I tried to let $x = \tan(u)$. Then
$$ \arctan(\tan(u)) + \arctan(\tan(\frac{\pi}{2} - x)) = \frac{\pi}{2}$$
but it does not seem useful.
I'd appreciate most a proof that gives intuition and / or uses geometric insight.
Inasmuch as the OP requested an approach that is intuitive or geometric, let's proceed accordingly. The ensuing heuristic discussion serves only to supplement the more analytical/rigorous approaches, and can help solidify the understanding of the relationship.
Suppose we have a right triangle formed from the three coordinate points $(0,0)$, $(1,0)$, and $(1,x)$. Note that the tangent of angle $\theta$ between the hypotenuse and the $x$ axis is
$$\tan \theta =x \tag 1$$
We also can see that the tangent of the opposite side angle $\phi$ is
$$\tan \phi = 1/x\tag 2$$
But we know that the sum of the angles $\theta$ and $\phi$ must add to $\pi/2$. We have, therefore, from $(1)$ and $(2)$ that
$$\theta +\phi =\arctan(x)+\arctan(1/x)=\pi/2$$
While in this development, the angles were restricted to be between $0$ and $\pi/2$, we can adapt this same approach show that the relationship is indeed general for $x>0$. And we can also use this approach to show that for $x<0$, $\arctan(x)+\arctan(1/x)=-\pi/2$.
$$f\left( x \right) =\arctan { \left( x \right) +\arctan { \left( \frac { 1 }{ x } \right) } } $$ $$f^{ \prime }\left( x \right) =\frac { 1 }{ 1+{ x }^{ 2 } } -\frac { 1 }{ 1+{ x }^{ 2 } } =0$$ $$\\ f\left( x \right) =c\\ f=f\left( 1 \right) $$
Let $$\tan ^{-1}(x)=\theta\iff x=\tan \theta$$ Where, $0<\theta<\frac{\pi}{2}$
Now, we know $$\tan\left(\frac{\pi}{2}-\theta\right)=\cot\theta$$ $$\tan\left(\frac{\pi}{2}-\theta\right)=\frac{1}{\tan\theta}$$ Setting $\tan\theta=x$ $$\tan\left(\frac{\pi}{2}-\theta\right)=\frac{1}{x}$$ $$\frac{\pi}{2}-\theta=\tan^{-1}\left(\frac{1}{x}\right)$$ $$\theta+\tan^{-1}\left(\frac{1}{x}\right)=\frac{\pi}{2}$$ Setting value of $\theta$ $$\tan^{-1}\left(x\right)+\tan^{-1}\left(\frac{1}{x}\right)=\frac{\pi}{2}$$
Let $f(x)=\arctan(x)+\arctan(\frac1x)$. Then $f'(x)=\frac{1}{1+x^2}+\frac{1}{1+1/x^2}\frac{-1}{x^2}=\frac{1}{1+x^2}-\frac{1}{1+x^2}=0$ so $f(x)$ is constant for $x>0$. Then note that $f(1)=\frac{\pi}{2}$ and thus $f(x)=\frac{\pi}{2}$.
There is a discontinuity at $x=0$, so the derivative only makes sense for $x\neq 0$. When $x<0$, you can check that $f(-1)=-\frac{\pi}{2}$. $f(x)$ has the same derivative so for $x<0$, $f(x) -\frac{\pi}{2}$
$$\arctan x + \arctan \frac{1}{x} = - \frac{\pi}{2}$$
– Tolaso Aug 20 '15 at 05:33$\arctan(a)+\arctan(b) =\arctan(\frac{a+b}{1-ab}) $.
Therefore $\arctan(x)+\arctan(1/x) =\arctan(\frac{x+1/x}{1-1}) =\arctan(\frac{x+1/x}{0}) =\pi/2 $.
If this bothers you,
$\begin{array}\\ \arctan(x-c)+\arctan(1/x) &=\arctan(\frac{x-c+1/x}{1-(x-c)/x)})\\ &=\arctan(\frac{x-c+1/x}{c/x})\\ &=\arctan(\frac{x^2-cx+1}{c})\\ &=\arctan(\frac{x^2+1}{c}-x)\\ &\to \pi/2 \quad\text{ as } c \to 0\\ \end{array} $.
Letting $x=\tan u$, we have $$\begin{array}{lll} \arctan(x)+\arctan(1/x)&=&\arctan(\tan u)+\arctan(1/\tan u)\\ &=&\arctan(\tan u)+\arctan(\cot u)\\ &=&\color{blue}{\arctan(\tan u)+\arctan(\tan (\frac{\pi}{2}-u))}\\ &=&u+\frac{\pi}{2}-u\\ &=&\frac{\pi}{2} \end{array}$$ The line highlighted in blue should be similar to your equation (note the difference though).
