If $\sqrt{1+\sec A}=x$, then the value of $x$ is:

Let's convert the expression inside the square root into terms of sine and cosine:
$$1+\sec A=1+\frac{1}{\cos A}=\frac{\cos A+1}{\cos A}$$
Now, let's look at the expression inside the square root of Option B:
$$\frac{\sin A\tan A}{1-\cos A}=\frac{\sin A\cdot \frac{\sin A}{\cos A}}{1-\cos A}=\frac{{\sin }^{2}A}{\cos A(1-\cos A)}$$
Using the identity ${\sin }^{2}A=1-{\cos }^{2}A=(1-\cos A)(1+\cos A)$:
$$\frac{(1-\cos A)(1+\cos A)}{\cos A(1-\cos A)}=\frac{1+\cos A}{\cos A}=1+\sec A$$
Thus, we have:
$$1+\sec A=\frac{\sin A\tan A}{1-\cos A}$$
Taking the square root on both sides:
$$\sqrt{1+\sec A}=\sqrt{\frac{\sin A\tan A}{1-\cos A}}$$
Therefore, $x=\sqrt{\frac{\sin A\tan A}{1-\cos A}}$.