Step 1: Understanding the Question:
We need to integrate the product of two cosecant functions with different arguments.
Step 2: Key Formula or Approach:
For integrals of the form $\int \csc(x-a)\csc(x-b) dx$, the standard technique is to multiply and divide the entire integral by $\sin(a-b)$. Here, the difference is simply $a$.
Step 3: Detailed Explanation:
The integral is $I = \int \frac{1}{\sin(x-a) \sin x} dx$.
Multiply and divide the expression by $\sin a$:
$$I = \frac{1}{\sin a} \int \frac{\sin a}{\sin(x-a) \sin x} dx$$
Rewrite the numerator $a$ as $(x) - (x-a)$:
$$I = \operatorname{cosec} a \int \frac{\sin(x - (x-a))}{\sin(x-a) \sin x} dx$$
Apply the compound angle formula $\sin(A-B) = \sin A \cos B - \cos A \sin B$:
$$I = \operatorname{cosec} a \int \frac{\sin x \cos(x-a) - \cos x \sin(x-a)}{\sin(x-a) \sin x} dx$$
Separate the fraction into two parts:
$$I = \operatorname{cosec} a \int \left[ \frac{\cos(x-a)}{\sin(x-a)} - \frac{\cos x}{\sin x} \right] dx$$
$$I = \operatorname{cosec} a \int [ \cot(x-a) - \cot x ] dx$$
Integrate using $\int \cot \theta d\theta = \log|\sin \theta|$:
$$I = \operatorname{cosec} a [ \log|\sin(x-a)| - \log|\sin x| ] + c$$
Apply the logarithm quotient rule $\log A - \log B = \log(A/B)$:
$$I = \operatorname{cosec} a \cdot \log\left| \frac{\sin(x-a)}{\sin x} \right| + c$$
Rewrite $\frac{1}{\sin x}$ as $\operatorname{cosec} x$:
$$I = \operatorname{cosec} a \cdot \log| \sin(x-a) \operatorname{cosec} x | + c$$
Step 4: Final Answer:
The evaluated integral matches option (A).