A

Given integrals to solve:
(i) \( \int \left( \frac{\tan x}{\cos x} \right) dx \)
(ii) \( \int \left( \frac{\cos x}{\sin^2 x} \right) dx \)

Step 1: Solve the first integral (i)
We can simplify \( \frac{\tan x}{\cos x} \) as follows:
\( \tan x = \frac{\sin x}{\cos x} \) so it becomes \( \frac{\sin x}{\cos^2 x} \).
Thus, the integral can be rewritten as:
\[ \int \frac{\sin x}{\cos^2 x} dx \]

Now we will use the substitution: \( u = \cos x \) which gives us \( du = -\sin x \, dx \).
Hence, we can rewrite the integral as:
\[ -\int \frac{1}{u^2} du = -\left( -\frac{1}{u} \right) + C = \frac{1}{\cos x} + C = \sec x + C \]

Step 2: Solve the second integral (ii)
For the second integral we need to evaluate:
\[ \int \left( \frac{\cos x}{\sin^2 x} \right) dx \]
We will rewrite this as:
\[ \int \cot x \csc x \, dx \]

For this integral, we can use the identity:
\( \frac{d}{dx} (\cot x) = -\csc^2 x \)
Therefore, we can rewrite this integral as:
\[ -\int d(\cot x) = -\cot x + C \]

Final Answers:
The solution to the first integral is \( \sec x + C \)
The solution to the second integral is \( -\cot x + C \). Hence, the correct answer corresponds to the first integral option A.