Cho \(\tan x = \dfrac{1}{2},\tan y = \dfrac{1}{3}\) với \(x,y \in \left( {0;\dfrac{\pi }{2}} \right)\) . Khi đó \(x + y\) bằng

* Hướng dẫn giải

Ta có: \(\tan \left( {x + y} \right) = \dfrac{{{\mathop{\rm tanx}\nolimits}  + \tan y}}{{1 - {\mathop{\rm tanx}\nolimits} {\mathop{\rm tany}\nolimits} }}\)\(\, = \dfrac{{\dfrac{1}{2} + \dfrac{1}{3}}}{{1 - \dfrac{1}{2}.\dfrac{1}{3}}} = \dfrac{{\dfrac{5}{6}}}{{\dfrac{5}{6}}} = 1\).

Do \(x,y \in \left( {0;\dfrac{\pi }{2}} \right)\) nên \(0 < x + y < \pi \). Suy ra \(x + y = \dfrac{\pi }{4}\).