We begin with the L.H.S. of the given equation:

$$\frac{\cos (\pi + x) \cos (-x)}{\sin (\pi - x) \cos \left( \frac{\pi}{2} + x \right)}$$

Using the standard trigonometric identities, we simplify the numerator terms. We know that $\cos (\pi + x) = -\cos x$. To evaluate $\cos (-x)$, we use the identity $\cos (x - y) = \cos x \cos y + \sin x \sin y$. Substituting $x=0$, we get:

$$\cos (0 - x) = \cos 0 \cos x + \sin 0 \sin x = \cos x$$

Thus, $\cos (-x) = \cos x$.

Next, we simplify the denominator terms. From the identities, we have $\sin (\pi - x) = \sin x$ and $\cos \left( \frac{\pi}{2} + x \right) = -\sin x$. Substituting these simplified values back into the expression, we get:

$$\frac{(-\cos x) \cdot (\cos x)}{(\sin x) \cdot (-\sin x)}$$

Simplifying the signs and terms, we have:

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

We know from the definition of cotangent that $\cot x = \frac{\cos x}{\sin x}$. Therefore, the expression becomes:

$$\left( \frac{\cos x}{\sin x} \right)^2 = \cot^2 x$$

Hence, L.H.S. = R.H.S., which proves the required identity.