因為 $\dfrac{\pi}{2} < x < \pi$，所以 $x$ 為第二象限角。又 $\sin x = \dfrac{3}{5}$，由廣義角三角函數定義與同角三角關係可得： $$\cos x = -\sqrt{1 - \sin^2 x} = -\sqrt{1 - \left(\dfrac{3}{5}\right)^2} = -\dfrac{4}{5}$$ $$\tan x = \dfrac{\sin x}{\cos x} = -\dfrac{3}{4}$$ $$\cot x = \dfrac{1}{\tan x} = -\dfrac{4}{3}$$ $$\sec x = \dfrac{1}{\cos x} = -\dfrac{5}{4}$$ $$\csc x = \dfrac{1}{\sin x} = \dfrac{5}{3}$$ 故正確選項為 $(3)(4)(5)$。