Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
1  
Do you mean $\arctan g(x)+\arctan g(\frac 1 x)=\pi /2?$ –  learner Aug 21 '13 at 11:55
2  
@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. –  Jyrki Lahtonen Aug 21 '13 at 11:58
1  
@JyrkiLahtonen Thank you sir for explanation. I was not aware of the notion. –  learner 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. –  Jyrki Lahtonen 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. –  amWhy Aug 21 '13 at 12:24

4 Answers 4

up vote 11 down vote accepted

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

share|improve this answer
    
This is by far the most visual solution, thank you. –  András Hummer 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.

share|improve this answer
2  
So $f$ is constant on each connected components of $\mathbb{R}^*$! ($f(x)=- \pi/2$ when $x<0$.) –  Seirios 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.

share|improve this answer

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

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.