Let $$A = \{\theta : \sin(\theta) = \tan(\theta)\}$$ and $$B = \{\theta : \cos(\theta) = 1\}$$ be two sets. Then :

First, we need to understand the sets A and B.

Set A is defined as $$ A = \{\theta : \sin(\theta) = \tan(\theta)\} $$.

Recall that $$ \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} $$, so we substitute this into the equation:

$$ \sin(\theta) = \frac{\sin(\theta)}{\cos(\theta)} $$

Note that $$ \tan(\theta) $$ is undefined when $$ \cos(\theta) = 0 $$, so we must exclude these points. Therefore, $$ \cos(\theta) \neq 0 $$.

Multiply both sides by $$ \cos(\theta) $$ (since it is not zero):

$$ \sin(\theta) \cdot \cos(\theta) = \sin(\theta) $$

Bring all terms to one side:

$$ \sin(\theta) \cdot \cos(\theta) - \sin(\theta) = 0 $$

Factor out $$ \sin(\theta) $$:

$$ \sin(\theta) (\cos(\theta) - 1) = 0 $$

This equation holds if either factor is zero:

$$ \sin(\theta) = 0 \quad \text{or} \quad \cos(\theta) - 1 = 0 $$

Which simplifies to:

$$ \sin(\theta) = 0 \quad \text{or} \quad \cos(\theta) = 1 $$

Now, we must ensure that these solutions satisfy $$ \cos(\theta) \neq 0 $$.

Case 1: $$ \sin(\theta) = 0 $$.
The solutions are $$ \theta = n\pi $$, where $$ n $$ is any integer.
At these points, $$ \cos(\theta) = \cos(n\pi) = (-1)^n $$, which is either 1 or -1, never zero. So these are valid.

Case 2: $$ \cos(\theta) = 1 $$.
The solutions are $$ \theta = 2m\pi $$, where $$ m $$ is any integer.
At these points, $$ \cos(\theta) = 1 \neq 0 $$, so they are valid.

Notice that when $$ \cos(\theta) = 1 $$, $$ \theta = 2m\pi $$, which is included in $$ \theta = n\pi $$ (by letting $$ n = 2m $$). Also, $$ \sin(2m\pi) = 0 $$, so the condition $$ \sin(\theta) = 0 $$ covers both cases. Therefore, the complete solution set for A is:

$$ A = \{\theta : \theta = n\pi, n \in \mathbb{Z}\} $$

Now, set B is defined as $$ B = \{\theta : \cos(\theta) = 1\} $$. The solutions are:

$$ B = \{\theta : \theta = 2m\pi, m \in \mathbb{Z}\} $$

We compare A and B.

First, note that every element of B is in A: if $$ \theta = 2m\pi $$, then $$ \theta = n\pi $$ with $$ n = 2m $$ (an even integer), so $$ B \subset A $$.

However, A contains additional points not in B. For example, take $$ \theta = \pi $$:
- $$ \pi \in A $$ because $$ \sin(\pi) = 0 $$ and $$ \tan(\pi) = 0 $$, so $$ 0 = 0 $$.
- But $$ \cos(\pi) = -1 \neq 1 $$, so $$ \pi \notin B $$.

Similarly, $$ \theta = -\pi, 3\pi, $$ etc., are in A but not in B.

Therefore, A is not a subset of B because there are elements in A that are not in B.

Now, examine the options:

A. A = B → False, because A has more elements (like $$ \pi $$) than B.

B. $$ A \not\subset B $$ → True, because A is not a subset of B (as shown with $$ \theta = \pi $$).

C. $$ B \not\subset A $$ → False, because B is a subset of A (as established).

D. $$ A \subset B $$ and $$ B - A \neq \phi $$ → False, because $$ A \subset B $$ is false (as above), and even if we consider the second part, $$ B - A $$ is empty since $$ B \subset A $$, so $$ B - A = \phi $$.

Hence, the correct answer is Option B.