Since $\sin\theta = \dfrac{1}{9}$, we have $\csc\theta = \dfrac{1}{\sin\theta} = 9$.
$$\frac{9\csc\theta + 1}{9\csc\theta - 1} = \frac{9(9) + 1}{9(9) - 1} = \frac{81 + 1}{81 - 1} = \frac{82}{80}$$
Answer: (D) $\dfrac{82}{80}$
Source: Chapter 8, Section 8.2 (Trigonometric Ratios)
The key step is recognising that $\csc\theta = \dfrac{1}{\sin\theta} = \dfrac{1}{1/9} = 9$. Substituting directly gives the value. Examiners expect students to recall the reciprocal relation $\csc\theta = \dfrac{1}{\sin\theta}$ instantly and substitute without complicated working.