TITLE: How can I prove that $\operatorname{arctg}(x) + \operatorname{arctg}(\frac{1}{x}) = \frac{\pi}{2}$, given that $x > 0$?
QUESTION [5 upvotes]: Which would be the easier way to prove that $\operatorname{arctg}(x) + \operatorname{arctg}(\frac{1}{x}) = \frac{\pi}{2}$ in cases where $x > 0$? I don't need explicit solutions, rather keywords and pointers towards the direction of a feasible method that can help me prove this equality.

REPLY [12 votes]: Draw a right-angled triangle with the two shorter sides having length $1$ and $x$ respectively. Adding the angles we get the equality $$\arctan(x)+\arctan \left(\frac{1}{x}\right)+\frac{\pi}{2}=\pi.$$

REPLY [3 votes]: We have $\tan (\arctan x)=x$ and also $\tan (\arctan \frac1x)=\frac1x \iff\cot (\arctan \frac1x)=x \iff \tan(\frac\pi2 - \arctan\frac1x)=x$.
Now try to combine $\tan (\arctan x)=x$ and $\tan(\frac\pi2 - \arctan\frac1x)=x$.