If $p=\frac{\sin A}{1+\cos A}$, then $\frac{\sin A}{1-\cos A}$ is equal to:

The correct answer is $\frac{1}{p}$.

Let us find the value of the given expression step-by-step.
We are given that:
$p=\frac{\sin A}{1+\cos A}$

We want to evaluate:
$X=\frac{\sin A}{1-\cos A}$

Let us multiply the numerator and the denominator of $X$ by $(1+\cos A)$:
$X=\frac{\sin A(1+\cos A)}{(1-\cos A)(1+\cos A)}$

Using the algebraic identity $(a-b)(a+b)=a^2-b^2$, the denominator becomes:
$(1-\cos A)(1+\cos A)=1-{\cos }^{2}A$

We know from the fundamental trigonometric identity that:
$1-{\cos }^{2}A={\sin }^{2}A$

Substituting this back into the expression for $X$, we get:
$X=\frac{\sin A(1+\cos A)}{{\sin }^{2}A}$

Canceling one factor of $\sin A$ from both the numerator and the denominator, we obtain:
$X=\frac{1+\cos A}{\sin A}$

Now, let us compare this result with the given expression for $p$:
$p=\frac{\sin A}{1+\cos A}$

It is clear that:
$X=\frac{1}{p}$

Therefore, $\frac{\sin A}{1-\cos A}$ is equal to $\frac{1}{p}$.