5.) $$ \frac{\cos \left(90^{\circ}-2 x\right)}{2-2 \sin (-x) \cdot \cos \left(90^{\circ}+x\right)} $$

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To simplify the expression:

$$
\frac{\cos \left(90^{\circ}-2 x\right)}{2-2 \sin (-x) \cdot \cos \left(90^{\circ}+x\right)}
$$

let's break it down step by step.

### Step 1: Simplify the Numerator
$$
\cos \left(90^{\circ} - 2x\right)
$$
Using the co-function identity:
$$
\cos(90^{\circ} - \theta) = \sin \theta
$$
So,
$$
\cos \left(90^{\circ} - 2x\right) = \sin(2x)
$$
And using the double-angle identity:
$$
\sin(2x) = 2 \sin x \cos x
$$

### Step 2: Simplify the Denominator
$$
2 - 2 \sin (-x) \cdot \cos \left(90^{\circ} + x\right)
$$
First, recall that:
$$
\sin(-x) = -\sin x
$$
$$
\cos(90^{\circ} + x) = -\sin x
$$
Substituting these:
$$
2 - 2 (-\sin x) \cdot (-\sin x) = 2 - 2 \sin^2 x
$$
Factor out the 2:
$$
2(1 - \sin^2 x)
$$
Using the Pythagorean identity:
$$
1 - \sin^2 x = \cos^2 x
$$
So,
$$
2 \cos^2 x
$$

### Step 3: Combine Numerator and Denominator
Now, substitute the simplified numerator and denominator back into the original expression:
$$
\frac{2 \sin x \cos x}{2 \cos^2 x}
$$
Cancel out the common factors:
$$
\frac{\sin x}{\cos x} = \tan x
$$

### Final Simplified Form
$$
\tan x
$$

**Answer:**  
After simplifying, the expression is equal to tangent x.

$$\tan x$$
The expression simplifies to $$\tan x$$.

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