Let $x$ be a positive number and let $\Theta =\tan^{-1}x$
Show that $\tan^{-1}\frac1x = \frac \pi2 - \Theta$ and hence show that $\tan^{-1}x + \tan^{-1}\frac1x = \frac\pi2$ for $x > 0$
Find $\tan^{-1}x + \tan^{-1}\frac1x$ for $x<0$
(I know you have to use the fact that $\tan^{-1}x$ is odd)
I know this is significant somewhere.
Where do I go from here?
The following is just one possible approach. Let $f : (0, \infty) \to \Bbb R$ be $f(x) = \arctan x + \arctan \frac 1 x$. Notice that
$$f'(x) = \frac 1 {1+x^2} + \frac 1 {1 + \frac 1 {x^2}} \left( - \frac 1 {x^2} \right) = \frac 1 {1+x^2} - \frac 1 {1+x^2} = 0 ,$$
which means that $f$ is constant. It remains to find out this constant value. In particular, it will be equal to $f(1)$ (chosen among others because it is easiest to evaluate). But
$$f(1) = \arctan 1 + \arctan \frac 1 1 = \arctan 1 + \arctan 1 = \frac \pi 4 + \frac \pi 4 = \frac \pi 2 ,$$
so $f(x) = f(1) = \frac \pi 2$.
Values other than $1$ that would have made evaluation easy would have been $\sqrt 3$ and $\frac 1 {\sqrt 3}$.
Alternatively, if you want to avoid derivatives, you could proceed using the trigonometric difference formula:
$$\qquad \tan(a-b) = \frac {\tan a - \tan b} {1 + \tan a \tan b}$$
which implies that
$$\tan \left(\frac \pi 2 - a \right) = \tan \left( \frac \pi 4 + \frac \pi 4 - a \right) = \frac {\tan \frac \pi 4 + \tan \left( \frac \pi 4 - a \right)} {1 - \tan \frac \pi 4 \cdot \tan \left( \frac \pi 4 - a \right)} = \frac {1 + \tan \left( \frac \pi 4 - a \right)} {1 - \tan \left( \frac \pi 4 - a \right)} = \\ \frac {1 + \frac {\tan \frac \pi 4 - \tan a} {1 + \tan \frac \pi 4 \cdot \tan a}} {1 - \frac {\tan \frac \pi 4 - \tan a} {1 + \tan \frac \pi 4 \cdot \tan a}} = \frac {1 + \frac {1 - \tan a} {1 + \tan a}} {1 - \frac {1 - \tan a} {1 + \tan a}} = \frac 2 {2 \tan a} = \frac 1 {\tan a} .$$
Rewriting your equality as $\arctan \frac 1 x = \frac \pi 2 - \arctan x$ and applying $\tan$ to both sides (because $\tan : (0, \frac \pi 2) \to (0, \infty)$ is bijective) you get
$$\frac 1 x = \tan \left( \arctan \frac 1 x \right) = \tan \left( \frac \pi 2 - \arctan x \right) = \frac 1 {\tan \arctan x} = \frac 1 x$$
which is a true equality, therefore the initial equality is also true.
To derive a similar formula for $x<0$ just remember that $\tan(-x) = -\tan x$ and use the identity that you already have.
Since $x>0$, you have $0<\theta<\pi/2$ and $x=\tan\theta$.
Then $$ \tan\left(\frac{\pi}{2}-\theta\right)=\cot\theta=\frac{1}{\tan\theta}=\frac{1}{x} $$
You can conclude.
If $x<0$, then $\arctan x=-\arctan(-x)$. Set $y=-x$ and use the former identity.
Alex's answer is correct. But there is a formal easy way. Let $x=\tan\theta$. If $x>0$ then we should consider $\theta>0$ and $$\cot\theta=\frac 1{\tan\theta}=\frac 1x$$ You may also know that $\cot\theta=\tan(\frac{\pi}2-\theta)$. Thus: $$\begin{align} \tan^{-1}x+\tan^{-1}\frac 1x&=\tan^{-1}\tan\theta+\tan^{-1}\cot\theta\\ &=\theta+\tan^{-1}\tan(\frac{\pi}2-\theta)\\&=\theta+\frac{\pi}2-\theta=\frac{\pi}2 \end{align}$$ You may also want to check out this question. As a side-note, in your question you stated that $$\tan(\frac{\pi}2-\theta)=\frac{1-\tan\theta}{1+\tan\theta}$$ which is wrong and this equality holds for $\tan(\frac{\pi}4-\theta)$
$\pi$and $\tan^{-1}$ as$\tan^{-1}$etc. Take a look at the link. It's not that hard – polfosol Dec 29 '16 at 21:15