B

Step 1: We will solve the integrals one by one.

(i) The integral is:
$$\int \frac{\tan x}{\cos x} \, dx$$
This can be rewritten as:
$$\int \frac{\sin x}{\cos^2 x} \, dx$$
Letting $u = \cos x$, then $du = -\sin x \, dx$. Substituting gives:
$$ -\int \frac{1}{u^2} \, du = \frac{1}{u} + C = \frac{1}{\cos x} + C = \sec x + C$$

(ii) The integral is:
$$\int \frac{\cos x}{\sin^2 x} \, dx$$
We can rewrite it as:
$$\int \cot x \csc x \, dx$$
This integrates to:
$$-\csc x + C$$

Final Answers:
(i) $\sec x + C$ and (ii) $-\csc x + C$. Since both results are derived correctly, the correct answer is option B.