We are given that $\tan x = 1$, where $0^\circ \le x \le 90^\circ$. We are asked to evaluate $\frac{1-\sin^2 x}{\cos x}$.

1. Problem Description

We are given that $\tan x = 1$, where $0^\circ \le x \le 90^\circ$. We are asked to evaluate $\frac{1-\sin^2 x}{\cos x}$.

2. Solution Steps

Since $\tan x = 1$ and $0^\circ \le x \le 90^\circ$, we know that $x = 45^\circ$.

We need to find $\sin x$ and $\cos x$.

$\sin 45^\circ = \frac{\sqrt{2}}{2}$ and $\cos 45^\circ = \frac{\sqrt{2}}{2}$.

Now, we can substitute these values into the expression $\frac{1 - \sin^2 x}{\cos x}$.

$\frac{1 - \sin^2 x}{\cos x} = \frac{1 - (\frac{\sqrt{2}}{2})^2}{\frac{\sqrt{2}}{2}} = \frac{1 - \frac{2}{4}}{\frac{\sqrt{2}}{2}} = \frac{1 - \frac{1}{2}}{\frac{\sqrt{2}}{2}} = \frac{\frac{1}{2}}{\frac{\sqrt{2}}{2}} = \frac{1}{2} \cdot \frac{2}{\sqrt{2}} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$.

Alternatively, we can use the trigonometric identity $\sin^2 x + \cos^2 x = 1$, which means $1 - \sin^2 x = \cos^2 x$.

Therefore, $\frac{1 - \sin^2 x}{\cos x} = \frac{\cos^2 x}{\cos x} = \cos x$.

Since $\tan x = 1$, we have $x = 45^\circ$.

Thus, $\cos x = \cos 45^\circ = \frac{\sqrt{2}}{2}$.

3. Final Answer

The final answer is $\frac{\sqrt{2}}{2}$ (Option C).