[tex]\text{arctg}\,x\text{ si }\text{arcctg}\, x \text{ sunt functii continue, deci, derivabile.}\\ \\ f(x) = \text{arctg}\, x+\text{arcctg}\, x\\ \\ f'(x) = \dfrac{1}{1+x^2}-\dfrac{1}{1+x^2}= 0\\ \\ \displaystyle f(x) = \int f'(x) \, dx = \int 0\, dx = 0+C \\ \\ f(0) =\text{arctg}\, 0 +\text{arcctg }0 = 0+\dfrac{\pi}{2}\\ f(0) = C \\ \\ \Rightarrow C = \dfrac{\pi}{2} \\ \\ \Rightarrow f(x) = \dfrac{\pi}{2} \Rightarrow \boxed{\text{arctg}\, x+\text{arcctg}\, x = \dfrac{\pi}{2}}[/tex]
