The correct answer is $\boxed{\text{B. cos c cos B}}$.
In a spherical triangle ABC right angled at C, the sine of the angle B is equal to the cosine of the angle C times the cosine of the angle B. This can be shown using the law of cosines for spherical triangles, which states that
$$c^2 = a^2 + b^2 – 2ab \cos C$$
where $c$ is the side opposite the right angle, $a$ and $b$ are the other two sides, and $C$ is the angle opposite side $c$.
In this case, we know that $C = 90^\circ$, so the law of cosines becomes
$$c^2 = a^2 + b^2 – 2ab$$
We can then take the sine of both sides to get
$$\sin c = \frac{\sin a}{\cos B}$$
or
$$\sin B = \cos c \cos B$$
Therefore, the sine of the angle B is equal to the cosine of the angle C times the cosine of the angle B.
The other options are incorrect because they do not take into account the fact that the triangle is right angled at C. For example, option A would be correct if the triangle were not right angled, but in this case it is not.