# How can I prove that $\operatorname{arctg}(x) + \operatorname{arctg}(\frac{1}{x}) = \frac{\pi}{2}$, given that $x > 0$?

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.

• Do you mean $\arctan g(x)+\arctan g(\frac 1 x)=\pi /2?$ Aug 21 '13 at 11:55
• @learner: No. He is asking about $$\arctan x+\arctan\frac1x.$$ In different parts of the world (and different sciences) different abbreviations are used. Somewhere $\operatorname{tg} x$ is the default way to write the tangent. Aug 21 '13 at 11:58
• @JyrkiLahtonen Thank you sir for explanation. I was not aware of the notion. Aug 21 '13 at 12:16
• No worries. But I'm not sure I fully approve of @amWhy's edit. This way there is less chance of confusion for sure. But some "originality" was lost. No biggie obviously. Aug 21 '13 at 12:20
• @Jyrki Apologies for loss of originality. I was simply using the "easy" route to get the font of the arctan/arctg function to present as an operator, rather than italicized, though surely, I could have used operatorname{arctg} x to do the same. Aug 21 '13 at 12:24

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

• This is by far the most visual solution, thank you. Aug 21 '13 at 12:10

Hint: $f'\equiv0\implies f\equiv C$ for some constant $C$.

In this case you only need to find one point $x_{0}$ s.t you can calculate $f(x_{0})$ to know what $f$ is.

• So $f$ is constant on each connected components of $\mathbb{R}^*$! ($f(x)=- \pi/2$ when $x<0$.) Aug 21 '13 at 12:09

Using the derivative as suggested by Belgi is one way. Another would be to combine the identities \begin{aligned} \tan x&=\frac1{\cot x}\\ \cot x&=\tan(\frac\pi2-x) \end{aligned} as well as the definition of $\arctan$ in a hopefully indicated way. In this case all the angles involved in the calculation will be in the first quadrant.

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 $\tan(\frac\pi2 - \arctan\frac1x)=x$.