Directions: Reduce the given trigonometric expression to its simplest form.

Which of the following expressions is equivalent to $\frac{1-\sin x\cos x}{{(\sin x-\cos x)}^2}$?

The correct option is $\frac{1-\sin x\cos x}{1-2\sin x\cos x}$.

To reduce the given trigonometric expression to its simplest form, consider the expression:

$\frac{1-\sin x\cos x}{{(\sin x-\cos x)}^2}$

First, expand the denominator using the algebraic expansion identity ${(a-b)}^2=a^2-2ab+b^2$:

${(\sin x-\cos x)}^2={\sin }^{2}x-2\sin x\cos x+{\cos }^{2}x$

Using the Pythagorean trigonometric identity ${\sin }^{2}x+{\cos }^{2}x=1$, substitute $1$ for ${\sin }^{2}x+{\cos }^{2}x$:

${(\sin x-\cos x)}^2=1-2\sin x\cos x$

Now, substitute the simplified denominator back into the original expression:

$\frac{1-\sin x\cos x}{{(\sin x-\cos x)}^2}=\frac{1-\sin x\cos x}{1-2\sin x\cos x}$

Hence, the simplified equivalent expression is $\frac{1-\sin x\cos x}{1-2\sin x\cos x}$.