Compute the value of $(\arctan \frac{1}{2}+\arctan \frac{1}{3})/(\operatorname{arccot} \frac{1}{2}+\operatorname{arccot} \frac{1}{3}) $

Question

I'm new to this site I came up with this question in my homeworks:

compute : $ \dfrac{\arctan \dfrac{1}{2}+\arctan \dfrac{1}{3}}{\operatorname{arccot} \dfrac{1}{2}+\operatorname{arccot} \dfrac{1}{3}} $

I don't know what idea can help here!What can i start with?

Thanks : )

Answer 1

Hello and welcome to the site!
$ \tan(\arctan(\frac{1}{2})+\arctan(\frac{1}{3}))=\frac{\tan(\arctan(\frac{1}{2}))+\tan(\arctan(\frac{1}{3}))}{1-\tan(\arctan(\frac{1}{2}))\tan(\arctan(\frac{1}{3}))} =\frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{2}\frac{1}{3}}=1 $
So $\arctan(\frac 12)+\arctan(\frac 13)=\frac {\pi}4$ And thus, $ \text{arccot}(\frac{1}{2})+\text{arccot}(\frac{1}{3})=\pi-\frac{\pi}{4}=\frac{3\pi}{4} $
So the final result will be $\frac{1}{3}$.
I hope this could help :)

Answer 2

Like showing $\arctan(\frac{2}{3}) = \frac{1}{2} \arctan(\frac{12}{5})$,

for $ab<1,$ $$\arctan a+\arctan b=\arctan\left(\frac{a+b}{1-ab}\right)$$

$$\arctan \frac12+\arctan \frac13=\arctan1=\cdots=\frac\pi4$$

and $$\arctan a+\text{arccot}\ a=\frac\pi2\implies\text{arccot}\ \frac12+\text{arccot}\ \frac13=\pi-\frac\pi4=\cdots$$