1. We need to evaluate

$$\\int \frac{\sin x}{\sin(x-\alpha)}\,dx$$

and compare it with

$$Ax + B\log \sin(x-\alpha) + C.$$

2. Express $\sin x$ in terms of $(x-\alpha)$:

$$\sin x = \sin\big((x-\alpha)+\alpha\big).$$

Using the identity

$$\sin(u+v)=\sin u\cos v+\cos u\sin v,$$

we get

$$\sin x = \sin(x-\alpha)\cos\alpha + \cos(x-\alpha)\sin\alpha.$$

3. Divide by $\sin(x-\alpha)$:

$$\frac{\sin x}{\sin(x-\alpha)} = \cos\alpha + \sin\alpha\,\frac{\cos(x-\alpha)}{\sin(x-\alpha)}.$$

So,

$$\frac{\sin x}{\sin(x-\alpha)} = \cos\alpha + \sin\alpha\,\cot(x-\alpha).$$

4. Integrate term by term:

$$\int \frac{\sin x}{\sin(x-\alpha)}\,dx = \int \cos\alpha\,dx + \sin\alpha\int \cot(x-\alpha)\,dx.$$

Since $\cos\alpha$ and $\sin\alpha$ are constants,

$$= x\cos\alpha + \sin\alpha\int \cot(x-\alpha)\,dx.$$

Now,

$$\int \cot(x-\alpha)\,dx = \log|\sin(x-\alpha)|.$$

Hence,

$$\int \frac{\sin x}{\sin(x-\alpha)}\,dx = x\cos\alpha + \sin\alpha\log|\sin(x-\alpha)| + C.$$

5. Comparing with

$$Ax + B\log \sin(x-\alpha) + C,$$

we obtain

$$A=\cos\alpha, \qquad B=\sin\alpha.$$

6. Therefore,

$$(A,B)=\left(\cos\alpha,\sin\alpha\right).$$

This matches option B.