Proof:

To prove the equation $\frac{\sin(A) - \cos(A)}{1} \div \frac{\sin(A) \cdot \cos(A) - 1}{1} = \frac{1}{\sec(A) - \tan(A)}$, we can simplify both sides and show that they are equal.

LHS:
$\frac{\sin(A) - \cos(A)}{1} \div \frac{\sin(A) \cdot \cos(A) - 1}{1}$

To divide fractions, we can multiply by the reciprocal of the second fraction:

$\frac{\sin(A) - \cos(A)}{1} \cdot \frac{1}{\frac{\sin(A) \cdot \cos(A) - 1}{1}}$

Simplifying the denominator:

$\frac{\sin(A) - \cos(A)}{1} \cdot \frac{1}{\sin(A) \cdot \cos(A) - 1}$

Combining the fractions:

$\frac{\sin(A) - \cos(A)}{\sin(A) \cdot \cos(A) - 1}$

RHS:
$\frac{1}{\sec(A) - \tan(A)}$

Recall that $\sec(A) = \frac{1}{\cos(A)}$ and $\tan(A) = \frac{\sin(A)}{\cos(A)}$. Substituting these values:

$\frac{1}{\frac{1}{\cos(A)} - \frac{\sin(A)}{\cos(A)}}$

Simplifying the denominator:

$\frac{1}{\frac{1 - \sin(A)}{\cos(A)}}$

Recall that dividing by a fraction is the same as multiplying by its reciprocal:

$\frac{1}{\frac{1 - \sin(A)}{\cos(A)}} \cdot \frac{\cos(A)}{1}$

Combining the fractions:

$\frac{\cos(A)}{1 - \sin(A)}$

Conclusion:
We have shown that the left-hand side (LHS) is equal to $\frac{\sin(A) - \cos(A)}{\sin(A) \cdot \cos(A) - 1}$ and the right-hand side (RHS) is equal to $\frac{\cos(A)}{1 - \sin(A)}$. Since both sides simplify to the same expression, we can conclude that $\frac{\sin(A) - \cos(A)}{1} \div \frac{\sin(A) \cdot \cos(A) - 1}{1} = \frac{1}{\sec(A) - \tan(A)}$.