Sabendo que $\arcsin \left(\dfrac{3}{5}\right)~=~\arccos \left(\dfrac{4}{5}\right)$, temos:$$\sin x~=~\pm \sqrt{ \dfrac{1-\cos \left[\arccos \left(\dfrac{4}{5}\right)\right]}{2}~}$$$$\cos x~=~\pm \sqrt{ \dfrac{1+\cos \left[\arccos \left(\dfrac{4}{5}\right)\right]}{2}~}$$Assim$$\tg x~=~\pm \sqrt{ \dfrac{1-\cos \left[\arccos \left(\dfrac{4}{5}\right)\right]}{1+\cos \left[\arccos \left(\dfrac{4}{5}\right)\right]}~}~=~\pm \sqrt{ \dfrac{1-\dfrac{4}{5}}{1+\dfrac{4}{5}}~}~=~\pm ~\dfrac{1}{3}$$O máximo valor de $\tg x$ é $\dfrac{1}{3}$ .$$\bf{Alternativa~(B)}$$