Seja $x = \arcsin{\dfrac{a-1}{a+1}}$ , então: \begin{matrix} \sin{x} = {\dfrac{a-1}{a+1}} &,& \sin^2{x} + \cos^2{x} = 1 &\Rightarrow& \cos{x} = {\dfrac{2\sqrt{a}}{a+1}} &\therefore& \tan{x} = {\dfrac{a-1}{2\sqrt{a}}} \end{matrix}Já o valor da expressão solicitada, \begin{matrix} \tan{ \left[x + \arctan{\dfrac{1}{2\sqrt{a}}}\right]} &=&{\dfrac{ \tan{x} + \dfrac{1}{2\sqrt{a}}}{1\ - \ \tan{x} \cdot \dfrac{1}{2\sqrt{a}}}} &=& {\dfrac{\dfrac{a-1}{2\sqrt{a}} + \dfrac{1}{2\sqrt{a}}}{1\ - \ \dfrac{a-1}{2\sqrt{a}} \cdot \dfrac{1}{2\sqrt{a}}}} \end{matrix}Portanto,\begin{matrix} \tan{ \left[\arcsin{\dfrac{a-1}{a+1}} + \arctan{\dfrac{1}{2\sqrt{a}}} \right]} &=& {\dfrac{2a\sqrt{a}}{3a+1}} \end{matrix}\begin{matrix} Letra \ (C) \end{matrix}